Proving Social Choice Possible

Abstract

in this paper we give a more elaborate development of our algorithm for social choice which generalizes Condorcet’s method and satisfies Arrow’s five criteria and two axioms. An example is worked out for the case of m, number of alternatives, equals 5. The steps of the algorithm itself are precisely explicated. A proof is presented that the algorithm provides a social choice solution for all values of m when individual and social choices are expressed in terms of the R (preference or indifference) operator, and for all values of n, the number of voters. The demonstration of a Social Welfare Function (SWF) and the proof that it provides solutions in all cases i.e.for all values of m and n, proves that social choice is indeed possible.

As was shown in the companion paper, A Social Choice Algorithm, the key to proving social choice possible is the admission of ties as possible solutions. In the cases for which there are tie solutions, an additional criterion can be used to winnow the solution set. We introduce the concept of “digital utility” to choose among the various ties the one or ones that have the best “goodness of fit” with the voters’ preferences.

Introduction

In our previous paper, A Social Choice Algorithm, we presented an algorithm which generates solutions (social choices) which satisfy Arrow’s criteria and axioms. A social choice may consist of a set of tie elements and is based solely on binary, pairwise comparisons. If a candidate dies, the solution generated by considering the individual voting lists with the dead candidate’s name blotted out is entirely consistent with the solution generated by taking the original solution, blotting out the dead candidate’s name and recomputing the solution. These solutions are also entirely consistent with the Condorcet solution when such a solution exists. The algorithm considered here, therefore, is a generalization of the Condorcet solution.

In this paper we give a more elaborate and precise definition of the algorithm and then go on to prove that it provides solutions for all values of m and n where m is the number of alternatives and n is the number of voters.

An Example of the Algorithm for m = 5

We consider the following case: 

aRb,	aRc,	aTd, eRa
			bRc,	bRd,	eRb
				cRd,	cTe
				dRe

where xRy means x is prefered or indifferent to y and xTy means xRy and 
yRx are tied so the solution is {xRy, yRx} in accordance with Arrow’s (1951) Axiom 1. In connection with Axiom 1, Arrow states: “...Axiom 1 does not exclude th possibility of both xRy and yRx.”

We have the following stage 4 solutions which form a matrix:

Stage 4 Letter 	
Combinations	Stage 4 Winning Matrix

(1)	a,b,c,d	abcd	abdc	bc(a,d)	cb(a,d)	(a,d)bc	(a,d)cb a(b,c)d
(2)	a,b,c,e	eabc	eacb	ab(c,e)	ba(c,e)	(c,e)ab	(c,e)ba e(a,b)c
(3)	a,b,d,e	bdea	deab	bade	eabd	abed	eb(a,d) (a,d)eb
		e(a,d)b	(a,b,d)e	b(a,d,e)
(4)	a,c,d,e	acde	eacd	c(a,d)e	(a,d)ec	da(c,e)	d(c,e)a e(a,d)c
		(c,e)da	(c,d)ea	c(a,d,e)	e(a,c,d)	(a,c,e)d
(5)	b,c,d,e	bcde	debc	cdbe	b(c,e)d	bd(c,e)	d(c,e)b				eb(c,d)	(c,e)bd

1) Consider Letter Combination (1). For each element in the winning set, insert an e in every possible position and compute the rating of the number of stage 4 winners covered as follows:

			Potential Element of 	
Stage 4 Element		Stage 5 Winning Set	Rating
						
	abcd		abcde		3
			abced		1
			abecd		1
			aebcd		1
			eabcd		2

2) Pick highest rated one: abcde. Update list of covered elements.

Covering Element		Covered Elements	How many times (cumulative)?

	abcde			abcd					1
			acde					1
				bcde					1

3) Check to see that no element is covered more than twice. If it is, don't add covering element to winning set and go back to (1). If it is not, go to (4).

4) Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once.

(1) Blot out an e: We get abcd. abcd in winning set.
(2) Blot out a d: We get abce. abce not in winning set, but covered only once.
(3) Blot out a c: We get abde. abde not in winning set, but
covered only once.
(4) Blot out a b: We get acde. acde in winning set.
(5) Blot out an a: We get bcde. bcde in winning set.

Winning set is now {abcde }. Elements not in stage 4 winning matrix, but covered only once: abce, abde

5) Go back to (1) and proceed with next element.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	abdc				abdce				1
					abdec			1
					abedc			2
					aebdc			1
					eabdc			3

Pick highest rated one: eabdc. Update list of covered elements.

Covering Element		Covered Elements         How many times (cumulative)?

	eabdc			abcd				1
					acde				1
					bcde				1
					abdc				1
					eabc				1
			eabd				1
Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: {abcde, eabdc }

(1) Blot out an e: We get abcd, abdc. abcd, abdc in
winning set.
(2) Blot out a d: We get abce, eabc. eabc in winning set. abce not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd. eabd in winning set. abde not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc. acde in winning set. eadc not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc. bcde in winning set. ebdc not in winning set, but covered only once.

Winning set is now {abcde, eabdc }. Elements not in stage 4 winning matrix, but covered only once: abce, abde, eadc, ebdc.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	bc(a,d)		bc(a,d)e				3
			bc(a,d,e)			3
			bce(a,d)				1
			bec(a,d)				1
			ebc(a,d)				2

Pick highest rated ones: bc(a,d)e and bc(a,d,e). Update list of covered elements.

Covering Elements		Covered Elements         How many times (cumulative)?

	bc(a,d)e		 	abcd				1
	bc(a,d,e)			acde				1
					bcde				2
					abdc				1
					eabc				1
			eabd				1
					bc(a,d)			2
					c(a,d)e			1
					b(a,d,e)			1			
			c(a,d,e)		1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: {abcde, eabdc, bc(a,d)e, bc(a,d,e)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), bc(a,d) in winning set.
(2) Blot out a d: We get abce, eabc, bcae, bcae. abce, eabc in
winning set. bcae not in winning set and covered twice so we must strike bc(a,d,e) from potential winning set. We go back and revise the covered elements:

Covering Elements		Covered Elements         How many times (cumulative)?

	bc(a,d)e		 	abcd				1
					acde				1
					bcde				2
					abdc				1
					eabc				1
			eabd				1
					bc(a,d)			1
					c(a,d)e			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: {abcde, eabdc, bc(a,d)e}

(1) Blot out an e: We get abcd, abdc, bc(a,d)) in winning set.
(2) Blot out a d: We get abce, eabc, bcae. eabc in winning set. abce, bcae not in winning set but bcae now is covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e. eabd in winning set. abde and b(a,d)e not in winning set, but covered only once.
(4) Blot out a b: We get acde, eadc, c(a,d)e. acde, c(a,d)e in winning set. eadc not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde. bcde in
winning set. ebdc not in winning set, but covered only once.

Winning set is now { abcde, eabdc, bc(a,d)e }. Elements not in stage 4 winning matrix, but covered only once: abce, bcae, abde, b(a,d)e, eadc, ebdc.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	cb(a,d)		cb(a,d)e				2
			cb(a,d,e)			3
			cbe(a,d)				1
			ceb(a,d)				2
			ecb(a,d)				2

Pick highest rated one: cb(a,d,e). Update list of covered elements.

Covering Element		Covered Elements	How many times (cumulative)?

	cb(a,d,e)			abcd				1
					acde				1
					bcde				2
					abdc				1
					eabc				1
			eabd				1
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 

		{abcde, eabdc, bc(a,d)e, cb(a,d,e)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d). All are in winning set.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e). eabc in winning 	
set. abce, bcae, cb(a,e) not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e). eabd, b(a,d,e) 	
in winning set. abde, b(a,d)e not in winning set, but
covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e). acde,
c(a,d)e, c(a,d,e) in winning set. eadc not in winning set, but
covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e). bcde in winning 	
set. ebdc, cb(d,e) not in winning set, but covered
only once.

Winning set is now {abcde, eabdc, bc(a,d)e, cb(a,d,e)}. Elements not in stage 4 winning matrix, but covered only once: abce, bcae, cb(a,e), abde, b(a,d)e, eadc, ebdc, cb(d,e).

The next stage 4 winner in order is (a,d)bc.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(a,d)bc		(a,d)bce				1
			(a,d)bec				2
			(a,d)ebc				4
			(a,d,e)bc			1
			e(a,d)bc				4

Pick highest rated one(s): (a,d)ebc and e(a,d)bc. Update list of covered elements.

Covering Elements		Covered Elements         How many times (cumulative)?

	(a,d)ebc			abcd				1
	e(a,d)bc			acde				1
					bcde				2
					abdc				1
					eabc				2
			eabd				1
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
				e(a,d)b				1
				e(a,d)c				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
	{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc }

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc. All are in winning set.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc. eabc in winning set. abce, bcae, cb(a,e), aebc not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b. eabd, b(a,d,e), (a,d)eb, e(a,d)b in winning set. abde, b(a,d)e not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c. acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c in winning set. eadc not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc. bcde, debc in winning set. ebdc, cb(d,e), edbc not in winning set, but covered only once.

Winning set is now {abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc}. Elements not in stage 4 winning matrix, but covered only once: abce, bcae, cb(a,e), aebc, abde, b(a,d)e, eadc, ebdc, cb(d,e), edbc.

The next stage 4 winner in order is (a,d)cb.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(a,d)cb		(a,d)cbe				2
			(a,d)ceb				2
			(a,d)ecb				3
			(a,d,e)cb			1
			e(a,d)cb				4

Pick highest rated one: e(a,d)cb. Update list of covered elements.

Covering Element		Covered Elements         How many times (cumulative)?

	e(a,d)cb				abcd				1
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				1
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 

	{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb }

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb. All are in winning set.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb. eabc, eacb in winning set. abce, bcae, cb(a,e), aebc not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b. eabd, b(a,d,e), (a,d)eb, e(a,d)b in winning set. abde, b(a,d)e not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c. acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c in winning set. eadc not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb. bcde, debc in winning set. ebdc, cb(d,e), edbc, edcb not in winning set, but covered only once.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb}. 

Elements not in stage 4 winning matrix, but covered only once: 
abce, bcae, cb(a,e), aebc, abde, b(a,d)e, eadc, ebdc, cb(d,e), edbc, edcb

The next stage 4 winner in order is a(b,c)d.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	a(b,c)d		a(b,c)de				2
			a(b,c)ed				2
			a(b,c,e)d			1
			ae(b,c)d				1
			ea(b,c)d				3

Pick highest rated one: ea(b,c)d. Update list of covered elements.

Covering Element		Covered Elements         How many times (cumulative)?

	ea(b,c)d				abcd				1
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				1
					a(b,c)d			1
					eacd				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 

{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d }

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d. All are in winning set.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c). eabc, eacb in winning set. abce, bcae, cb(a,e), aebc, ea(b,c) not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd. eabd, b(a,d,e), (a,d)eb, e(a,d)b in winning set. abde, b(a,d)e not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd. acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, eacd in winning set. eadc not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d. bcde, debc in winning set. ebdc, cb(d,e), edbc, edcb, e(b,c)d not in winning set, but covered only once.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d }. 

Elements not in stage 4 winning matrix, but covered only once: 
abce, bcae, cb(a,e), aebc, ea(b,c), abde, b(a,d)e, eadc, ebdc, cb(d,e), edbc, edcb, e(b,c)d

The next stage 4 element in order that is not already covered twice is eacb. It is covered once already though. 

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	eacb		eacbd				3
			eacdb				2
			eadcb				1
			edacb				1
			deacb				2

The element with the highest rating is eacbd. However, eabd has already been covered twice so this element will not work. Let’s try eacdb.

Covering Element		Covered Elements         How many times (cumulative)?

	eacdb				abcd				1
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 

{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb }

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb. All are in winning set except acdb.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb. eabc, eacb in winning set. abce, bcae, cb(a,e), aebc, ea(b,c) not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb. eabd, b(a,d,e), (a,d)eb, e(a,d)b in winning set. abde, b(a,d)e, eadb not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd. acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, eacd, eacd in winning set. eadc not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb. bcde, debc in winning set. ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb not in winning set, but covered only once.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb }. 
Elements not in stage 4 winning matrix, but covered only once: 
acdb, abce, bcae, cb(a,e), aebc, ea(b,c), abde, b(a,d)e, eadb, eadc, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb

The next stage 4 element in order that is not already covered twice is ab(c,e).

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	ab(c,e)		ab(c,e)d				3
			ab(c,d,e)				1
			abd(c,e)				3
			adb(c,e)				2
			dab(c,e)				2

Pick highest rated one: ab(c,e)d. Update list of covered elements.

Covering Element		Covered Elements	How many times (cumulative)?

	ab(c,e)d				abcd				2
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				1
					b(c,e)d			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd. All are in winning set except acdb.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e). eabc, eacb, ab(c,e) in winning set. abce, bcae, cb(a,e), aebc, ea(b,c) not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed. eabd, b(a,d,e), (a,d)eb, e(a,d)b, abed in winning set. abde, b(a,d)e, eadb not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d. acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, eacd, eacd in winning set. eadc, a(c,e)d not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d. bcde, debc in winning set. ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb not in winning set, but covered only once.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d}. 
Elements not in stage 4 winning matrix, but covered only once: 
acdb, abce, bcae, cb(a,e), aebc, ea(b,c), abde, b(a,d)e, eadb, eadc, a(c,e)d, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb

abd(c,e) although rated 3 above is unuseable since abde has been covered once already.

The next stage 4 element in order that is not already covered twice is ba(c,e).

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	ba(c,e)		ba(c,e)d				2
			ba(c,d,e)			1
			bad(c,e)				3
			bda(c,e)				3
			dba(c,e)				2

Pick highest rated one: bad(c,e). Update list of covered elements.

Covering Element		Covered Elements	How many times (cumulative)?

	bad(c,e)				abcd				2
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				1
					b(c,e)d			1
					ba(c,e)			1
					bade				1
					bd(c,e)			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc. All are in winning set except acdb, badc.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e). eabc, eacb, ab(c,e), ba(c,e) in winning set. abce, bcae, cb(a,e), aebc, ea(b,c) not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade. eabd, b(a,d,e), (a,d)eb, e(a,d)b, abed, bade in winning set. abde, b(a,d)e, eadb not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e). acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, eacd, eacd in winning set. eadc, a(c,e)d, ad(c,e) not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e). bcde, debc, bd(c,e) in winning set. ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb not in winning set, but covered only once.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e)}. 
Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, abce, bcae, cb(a,e), aebc, ea(b,c), abde, b(a,d)e, eadb, eadc, a(c,e)d, ad(c,e), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, b(c,e)d

bda(c,e) is rated 3 above and is the next element considered.

Covering Element		Covered Elements	How many times (cumulative)?

	bda(c,e)				abcd				2
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				1
					bced				1
					ba(c,e)			2
					bade				1
					bd(c,e)			2
					da(c,e)			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac. All are in winning set except acdb, badc, bdac.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e). eabc, eacb, ab(c,e), ba(c,e), ba(c,e) in winning set. abce, bcae, cb(a,e), aebc, ea(b,c) not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae. eabd, b(a,d,e), (a,d)eb, e(a,d)b, abed, bade in winning set. abde, b(a,d)e, eadb, bdae not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e). acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, eacd, eacd, da(c,e) in winning set. eadc, a(c,e)d, ad(c,e) not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e). bcde, debc, bd(c,e), bd(c,e) in winning set. ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb not in winning set, but covered only once.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e)}. 
Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, abce, bcae, cb(a,e), aebc, ea(b,c), abde, b(a,d)e, eadb, bdae, eadc, a(c,e)d, ad(c,e), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb

The next stage 4 winner in order is (c,e)ab.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(c,e)ab		(c,e)abd				2
			(c,e)adb				1
			(c,e)dab				2
			(c,d,e)ab			1
			d(c,e)ab				4

Pick highest rated one: d(c,e)ab. Update list of covered elements.

Covering Element		Covered Elements	How many times (cumulative)?

	d(c,e)ab				abcd				2
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				1
					ba(c,e)			2
					bade				1
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				1
					d(c,e)a			1
					d(c,e)b			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab. All are in winning set except acdb, badc, bdac, dcab.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab. eabc, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab in winning set. abce, bcae, cb(a,e), aebc, ea(b,c) not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab. eabd, b(a,d,e), (a,d)eb, e(a,d)b, abed, bade, deab in winning set. abde, b(a,d)e, eadb, bdae not in winning set, but covered only once. 
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a. acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a in winning set. eadc, a(c,e)d, ad(c,e) not in winning set, but covered only once.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b. bcde, debc, bd(c,e), bd(c,e), d(c,e)b in winning set. ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb not in winning set, but covered only once.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab }. 
Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, abce, bcae, cb(a,e), aebc, ea(b,c), abde, b(a,d)e, eadb, bdae, eadc, a(c,e)d, ad(c,e), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb

The next stage 4 winner in order is (c,e)ba.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(c,e)ba		(c,e)bad				2
			(c,e)bda				3
			(c,e)dba				2
			(c,d,e)ba			1
			d(c,e)ba				3

Pick highest rated one: (c,e)bda. Update list of covered elements.

Covering Element		Covered Elements	How many times (cumulative)?

	(c,e)bda				abcd				2
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				1
					ba(c,e)			2
					bade				1
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				1
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda. All are in winning set except acdb, badc, bdac, dcab, cbda.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba. eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba in winning set. abce, bcae, cb(a,e), aebc, ea(b,c) not in winning set, but covered only once.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda. eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab in winning set. abde, b(a,d)e, eadb, bdae, ebda not in winning set, but covered only once.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da. acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da in winning set. eadc, a(c,e)d, ad(c,e) not in winning set, but covered only once.
(5) Blot out an a: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da. acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da in winning set. eadc, a(c,e)d, ad(c,e) not in winning set, but covered only once.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda}. 

Elements not in stage 4 winning matrix, but covered only once:
 acdb, badc, bdac, dcab, cbda, abce, bcae, cb(a,e), aebc, ea(b,c), abde, b(a,d)e, eadb, bdae, ebda, eadc, a(c,e)d, ad(c,e), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb

The next stage 4 winner in order is e(a,b)c.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	e(a,b)c		e(a,b)cd				2
			e(a,b)dc				1
			e(a,b,d)c			2
			ed(a,b)c				1
			de(a,b)c				2

e(a,b)cd and e(a,b,d)c don’t work since eacd and e(a,d)c have already been covered twice. Try de(a,b)c.

Covering Element		Covered Elements	How many times (cumulative)?

	de(a,b)c				abcd				2
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				1
					ba(c,e)			2
					bade				1
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				2
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c).
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b). In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, ea(,c)d, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, de(a,)c. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c}. 

Elements not in stage 4 winning matrix, but covered only once: acdb, badc, bdac, dcab, cbda, d(a,b)c, abce, bcae, cb(a,e), aebc, ea(b,c), abde, b(a,d)e, eadb, bdae, ebda, de(a,b), eadc, a(c,e)d, ad(c,e), deac, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb.

The next stage 4 winner in order is bdea.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	bdea		bdeac				1
			bdeca				1
			bdcea				1
			bcdea				2
			cbdea				1

bcdea doesn’t work since bcde has already been covered twice. Try bdeca.

Covering Element		Covered Elements	How many times (cumulative)?

	bdeca				abcd				2
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				1
					ba(c,e)			2
					bade				1
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				2
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, ea(,c)d, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, de(a,)c, deca. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca}. 

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, abce, bcae, cb(a,e), aebc, ea(b,c), beca, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), eadc, a(c,e)d, ad(c,e), deac, deca, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec.

The next stage 4 winner in order is deab. deab has already been covered twice. Next is bade.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	bade		badec				1
			badce				1
			bacde				3
			bcade				2
			cbade				1
			(b,c)ade				1

Try cbade.

Covering Element		Covered Elements	How many times (cumulative)?

	cbade				abcd				2
					acde				1
					bcde				2
					abdc				1
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				1
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				2
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, ea(,c)d, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, de(a,)c, deca, cade. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade}. 

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), eadc, a(c,e)d, ad(c,e), deac, deca, cade, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde

The next stage 4 winner in order is eabd. eabd has already been covered twice. Next is abed.

			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	abed		abedc				2
			abecd				2
			abced				2
			acbed				1
			cabed				1

Try abedc.

Covering Element		Covered Elements	How many times (cumulative)?

	abedc				abcd				2
					acde				1
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				2
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, ea(,c)d, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, de(a,)c, deca, cade, aedc. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, badec, abedc}. 

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc

The next stage 4 winner in order is eb(a,d).
 
			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	eb(a,d)		eb(a,d)c				2
			eb(a,c,d)			3
			ebc(a,d)				1
			ecb(a,d)				1
			ceb(a,d)				2

Try eb(a,c,d).

Covering Element		Covered Elements	How many times (cumulative)?

	eb(a,c,d)			abcd				2
					acde				1
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			1
					(a,d)ec			1
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			1
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				2
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d). In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c). In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c).
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d). In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d). Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, ea(,c)d, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, de(a,)c, deca, cade, aedc, e(a,c,d). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d). In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d)}. 

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), abde, b(a,d)e, eadb, bdae, ebda, de(a,b), eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc

The next stage 4 winner in order is (a,d)eb.
 
			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(a,d)eb		(a,d)ebc				4
			(a,d)ecb				3
			(a,d)ceb				2
			(a,c,d)eb			1
			c(a,d)eb				1

Try (a,d)ecb.

Covering Element		Covered Elements	How many times (cumulative)?

	(a,d)ecb				abcd				2
					acde				1
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			1
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				2
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb }

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, ea(,c)d, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, de(a,)c, deca, cade, aedc, e(a,c,d), (a,d)ec. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb}. 

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb

The next stage 4 winner in order is e(a,d)b. e(a,d)b has already been covered twice. Next try (a,b,d)e. 
 
			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(a,b,d)e		(a,b,d)ec			1
			(a,b,d)ce			1
			(a,b,c,d)e			1
			c(a,b,d)e			2

Try c(a,b,d)e.

Covering Element		Covered Elements	How many times (cumulative)?

	c(a,b,d)e			abcd				2
					acde				1
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			1
					c(a,d)e			2
					cb(a,d)			1
					b(a,d,e)			1
					c(a,d,e)			1
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				2
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1
					(a,b,d)e			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, c(a,b,d)e}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, c(a,b,d). In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), c(a,b,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, c(a,b)e. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, c(a,b)e.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, ea(,c)d, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, de(a,)c, deca, cade, aedc, e(a,c,d), (a,d)ec, c(a,d)e. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d)e. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, c(b,d)e. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, c(b,d)e.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, c(a,b,d)e}. 

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), c(a,b,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, c(a,b)e, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, c(b,d)e

The next stage 4 winner in order is b(a,d,e).
 
			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	b(a,d,e)		b(a,d,e)c			1
			b(a,c,d,e)			1
			bc(a,d,e)			3
			cb(a,d,e)			3

Try bc(a,d,e).

Covering Element		Covered Elements	How many times (cumulative)?

	bc(a,d,e)			abcd				2
					acde				1
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2
					cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				1
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				2
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1
					(a,b,d)e			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, c(a,b,d)e, bc(a,d,e)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, c(a,b,d), bc(a,d). In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), c(a,b,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, c(a,b)e, bc(a,e). In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, c(a,b)e, bc(a,e).
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e). In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e). Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, ea(,c)d, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, de(a,)c, deca, cade, aedc, e(a,c,d), (a,d)ec, c(a,d)e, c(a,d,e). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d)e, c(a,d,e). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, c(b,d)e, bc(d,e). In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, c(b,d)e, bc(d,e).

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, c(a,b,d)e, bc(a,d,e}. 

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), c(a,b,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, c(a,b)e, bc(a,e), abde, b(a,d)e, eadb, bdae, ebda, de(a,b), eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, c(b,d)e, bc(d,e)

The next stage 4 winner in order is acde.
 
			Potential Element of 	
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	acde		acdeb			1
			acdbe			2
			acbde			1
			abcde			3
			bacde			2
			acd(b,e)			1
			ac(b,d)e			1
			a(b,c)de			1
			(a,b)cde			1

All the above conflict in one way or other so we must go back and change a previous member of the winning set. We take out c(a,b,d)e which covers (a,b)de and c(a,d) e. We reduce the coverage of c(a,d)e by one. We take out the “left-overs” of c(a,b,d)e: c(a,b,d), c(a,b)e, c(b,d)e. We put in (a,b,c,d)e to cover (a,b,d)e and “left-overs”: (a,b,d,e), (a,b,c)e, (a,b,d)e, (b,c,d)e. Then we can use ac(b,d)e to cover acde.

Covering Element		Covered Elements	How many times (cumulative)?

	(a,b,c,d)e			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			1								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				1
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1
					(a,b,d)e			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d). In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e). Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e

The next stage 4 winner in order is eacd. eacd has already been covered twice. Next stage 4 winner is c(a,d)e.
 
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	c(a,d)e		c(a,d)eb				2
			c(a,d)be				2
			c(a,b,d)e			2
			cb(a,d)e				2
			bc(a,d)e				3

Try c(a,d)be.

Covering Element		Covered Elements	How many times (cumulative)?

	c(a,d)be				abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			1
					(c,e)ab			1
					deab				1
					d(c,e)a			1
					d(c,e)b			1
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1
					(a,b,d)e			1
					cdbe				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e). Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e
			
The next stage 4 winner in order is (a,d)ec. (a,d)ec has already been covered twice. Next stage 4 winner is da(c,e).
 
Next Stage 4 Winner	Stage 5 Winning Set	Rating

	da(c,e)		da(c,e)b				2
			da(b,c,e)			1
			dab(c,e)				2
			dba(c,e)				2
			bda(c,e)				3

Try da(c,e)b.

Covering Element		Covered Elements	How many times (cumulative)?

	da(c,e)b				abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			1
					deab				1
					d(c,e)a			1
					d(c,e)b			2
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1
					(a,b,d)e			1
					cdbe				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e). Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e
			
The next stage 4 winner in order is d(c,e)a.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	d(c,e)a		d(c,e)ab				4
			d(c,e)ba				3
			d(b,c,e)a			1
			db(c,e)a				1
			bd(c,e)a				3

Try d(b,c,e)a.

Covering Element		Covered Elements	How many times (cumulative)?

	d(b,c,e)a			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			1
					deab				1
					d(c,e)a			2
					d(c,e)b			2
					(c,e)bd			1
					(c,e)da			1
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1
					(a,b,d)e			1
					cdbe				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b, d(b,c,e)a}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb, d(b,c)a. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb, d(b,e)a. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e). Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b, d(b,c,e). In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e).

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e)		

 The next stage 4 winner in order is e(a,d)c. e(a,d)c has already been covered twice. The next stage 4 winner is (c,e)da.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(c,e)da		(c,e)dab				2
			(c,e)dba				2
			(c,e)bda				3
			(b,c,e)da			1
			b(c,e)da				2

Try (c,e)dab.

Covering Element		Covered Elements	How many times (cumulative)?

	(c,e)dab				abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				1
					d(c,e)a			2
					d(c,e)b			2
					(c,e)bd			1
					(c,e)da			2
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1
					(a,b,d)e			1
					cdbe				1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b, d(b,c,e)a, (c,e)dab}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, (c,e)ab. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e). Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b, d(b,c,e), (c,e)db. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db

The next stage 4 winner is (c,d)ea.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(c,d)ea		(c,d)eab				2
			(c,d)eba				1
			(c,d)bea				1
			(b,c,d)ea			1
			b(c,d)ea				2

Try (c,d)eab.

Covering Element		Covered Elements	How many times (cumulative)?

	(c,d)eab				abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			2
					(c,e)bd			1
					(c,e)da			2
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			1
					eb(c,d)			1
					(a,b,d)e			1
					cdbe				1
					(c,d)ea			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, (c,e)ab, ceab. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, deab. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b, d(b,c,e), (c,e)db, (c,d)eb. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb

The next stage 4 winner is c(a,d,e). c(a,d,e) has already been covered twice. The next stage 4 winner is e(a,c,d).

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	e(a,c,d)		e(a,c,d)b			1
			e(a,b,c,d)			1
			eb(a,c,d)			3
			be(a,c,d)			1

Try e(a,c,d)b.

Covering Element		Covered Elements	How many times (cumulative)?

	e(a,c,d)b			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			2
					(c,e)bd			1
					(c,e)da			2
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			1
					(a,b,d)e			1
					cdbe				1
					(c,d)ea			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, (c,e)ab, ceab, e(a,c)b. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, deab, e(a,d)b. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b

The next stage 4 winner is (a,c,e)d.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(a,c,e)d		(a,c,e)db			1
			(a,c,e)bd			2
			(a,b,c,e)d			1
			b(a,c,e)d			2

Try (a,c,e)bd.

Covering Element		Covered Elements	How many times (cumulative)?

	(a,c,e)bd			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			1
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			2
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			1
					(a,b,d)e			1
					cdbe				1
					(c,d)ea			1
					(a,c,e)d			1

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, (a,e)bd.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, (a,e)bd, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b

The next stage 4 winner is b(c,e)d.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	b(c,e)d		b(c,e)da				2
			b(c,e)ad				1
			b(a,c,e)d			2
			ba(c,e)d				2
			ab(c,e)d				4

Try b(a,c,e)d.

Covering Element		Covered Elements	How many times (cumulative)?

	b(a,c,e)d			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			2
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			2
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			1
					(a,b,d)e			1
					cdbe				2
					(c,d)ea			1
					(a,c,e)d			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e). In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e).
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd, dbea, b(a,e)d. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, cdea, (a,c,e)d. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, (a,c,e)d. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd, cdbe, b(c,e)d. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd, cdbe. Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d

The last stage 4 winner that hasn’t already been covered twice is eb(c,d).

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	eb(c,d)		eb(c,d)a				1
			eb(a,c,d)			3
			eba(c,d)				1
			eab(c,d)				1
			aeb(c,d)				1

Try eba(c,d).

Covering Element		Covered Elements	How many times (cumulative)?

	eba(c,d)				abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			1
					eacd				2
					ab(c,e)			1
					b(c,e)d			2
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			2
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			2
					(a,b,d)e			1
					cdbe				2
					(c,d)ea			1
					(a,c,e)d			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, eba(c,d)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d). In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, cdea, (a,c,e)d, ea(c,d). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, (a,c,e)d. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ea(c,d).
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd, cdbe, b(c,e)d, eb(c,d). In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd, cdbe, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, eba(c,d)}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ea(c,d), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d

Now all elements have been covered at least once. We go back and, starting at the beginning of the stage 4 winning matrix, cover again those elements that have only been covered once. The first such element is a(b,c)d.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	a(b,c)d		a(b,c)de				2
			a(b,c)ed				2
			a(b,c,e)d			1
			ae(b,c)d				1
			ea(b,c)d				3
			a(b,c)(d,e)			1
			a(b,c,e)d			1
			(a,e)(b,c)d			1
Try a(b,c)(d,e).

Covering Element		Covered Elements	How many times (cumulative)?

	a(b,c)(d,e)			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			2
					eacd				2
					ab(c,e)			1
					b(c,e)d			2
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			2
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			2
					(a,b,d)e			1
					cdbe				2
					(c,d)ea			1
					(a,c,e)d			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, eba(c,d), a(b,c)(d,e)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), a(b,c)d. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d), a(b,c)d. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e). In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, cdea, (a,c,e)d, ea(c,d), ac(d,e). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, da(c,e), d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, (a,c,e)d. Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ea(c,d), ac(d,e).
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(c,e)b, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd, cdbe, b(c,e)d, eb(c,d), (b,c)(d,e). In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd, cdbe, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e).

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, eba(c,d), a(b,c)(d,e)}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, dacb, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, a(c,e)b, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, daeb, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e), eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ea(c,d), ac(d,e), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e)

The next such element is ab(c,e).

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	ab(c,e)		ab(c,e)d				4
			ab(c,d,e)			1
			abd(c,e)				3
			adb(c,e)				1
			dab(c,e)				2
			a(b,d)(c,e)			1
			(a,d)b(c,e)			2
			
None of the 1-coverers work. Try dab(c,e). This means we must drop da(c,e)b from the winning set, find a 1-coverer for d(c,e)b and eliminate the by-products of da(c,e)b from the list of elements not in stage 4 matrix but covered only once. 

Covering Element		Covered Elements	How many times (cumulative)?

	dab(c,e)				abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			2
					eacd				2
					ab(c,e)			2
					b(c,e)d			2
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			1
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			1
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			2
					(a,b,d)e			1
					cdbe				2
					(c,d)ea			1
					(a,c,e)d			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, eba(c,d), a(b,c)(d,e), dab(c,e)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), a(b,c)d, dabc. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d), a(b,c)d. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), dabc.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e, ab(c,e). In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab, ab(c,e). Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e), dabe. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e), dabe.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, cdea, (a,c,e)d, ea(c,d), ac(d,e), da(c,e). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, (a,c,e)d, da(c,e). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ea(c,d), ac(d,e).
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd, cdbe, b(c,e)d, eb(c,d), (b,c)(d,e), db(c,e). In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd, cdbe, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e).

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, eba(c,d), a(b,c)(d,e), dab(c,e)}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), dabc, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e), dabe, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ea(c,d), ac(d,e), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e)

The next such element is (c,e)ba.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	(c,e)ba		(c,e)bad				2
			(c,e)bda				3
			(c,e)dba				2
			(c,d,e)ba			1
			d(c,e)ba				3
			(c,e)b(a,d)			3
			(c,e)(b,d)a			2
			
Try (c,d,e)ba.

Covering Element		Covered Elements	How many times (cumulative)?

	(c,d,e)ba			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			2
					eacd				2
					ab(c,e)			2
					b(c,e)d			2
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			1
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			2
					e(a,b)c			1
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			2
					(a,b,d)e			1
					cdbe				2
					(c,d)ea			1
					(a,c,e)d			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, eba(c,d), a(b,c)(d,e), dab(c,e), (c,d,e)ba}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), a(b,c)d, dabc, (c,d)ba. In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d), a(b,c)d. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), dabc, (c,d)ba.
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e, ab(c,e), (c,e)ba. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab, ab(c,e), (c,e)ba. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e), dabe, (d,e)ba. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e), dabe, (d,e)ba.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, cdea, (a,c,e)d, ea(c,d), ac(d,e), da(c,e), (c,d,e)a. In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, (a,c,e)d, da(c,e). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ea(c,d), ac(d,e), (c,d,e)a.
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd, cdbe, b(c,e)d, eb(c,d), (b,c)(d,e), db(c,e), (c,d,e)b. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd, cdbe, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, eba(c,d), a(b,c)(d,e), dab(c,e), (c,d,e)ba}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, ba(c,d), dabc, (c,d)ba, abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), ebac, a(b,c)e, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ebad, ab(d,e), dabe, (d,e)ba, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ea(c,d), ac(d,e), (c,d,e)a, ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b

The next such element is e(a,b)c.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	e(a,b)c		e(a,b)cd				2
			e(a,b)dc				1
			e(a,b,d)c			2
			ed(a,b)c				1
			de(a,b)c				2
			e(a,b)(c,d)			2
			(d,e)(a,b)c			1
			
None of the 1 or 2-coverers work. Try e(a,b)(c,d). This means we must drop eba(c,d) from the winning set since both e(a,b)(c,d) and eba(c,d) have the by-product ea(c,d) and they both cover eb(c,d). We must eliminate the by-products of eba(c,d) from the list of elements not in stage 4 matrix but covered only once. 

Covering Element		Covered Elements	How many times (cumulative)?

	e(a,b)(c,d)			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			1
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			2
					eacd				2
					ab(c,e)			2
					b(c,e)d			2
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			1
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			2
					e(a,b)c			2
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			2
					(a,b,d)e			1
					cdbe				2
					(c,d)ea			1
					(a,c,e)d			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, a(b,c)(d,e), dab(c,e), (c,d,e)ba, e(a,b)(c,d)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, a(b,c)d, dabc, (c,d)ba, (a,b)(c,d). In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d), a(b,c)d. Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, dabc, (c,d)ba, (a,b)(c,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e, ab(c,e), (c,e)ba, e(a,b)c. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab, ab(c,e), (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d. In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d.
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, cdea, (a,c,e)d, ac(d,e), da(c,e), (c,d,e)a, ea(c,d). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, (a,c,e)d, da(c,e). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ac(d,e), (c,d,e)a, ea(c,d).
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd, cdbe, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b, eb(c,d). In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd, cdbe, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, a(b,c)(d,e), dab(c,e), (c,d,e)ba, e(a,b)(c,d)}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, dabc, (c,d)ba, (a,b)(c,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d, eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ac(d,e), (c,d,e)a, ea(c,d), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b

The next such element is cb(a,d).

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	cb(a,d)		cb(a,d)e				3
			cb(a,d,e)			3
			cbe(a,d)				1
			ceb(a,d)				2
			ecb(a,d)				2
			c(b,e)(a,d)			1
			(c,e)b(a,d)			4
			
Try c(b,e)(a,d).

Covering Element		Covered Elements	How many times (cumulative)?

	c(b,e)(a,d)			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			2
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			2
					eacd				2
					ab(c,e)			2
					b(c,e)d			2
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			1
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			2
					e(a,b)c			2
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			2
					(a,b,d)e			1
					cdbe				2
					(c,d)ea			1
					(a,c,e)d			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, a(b,c)(d,e), dab(c,e), (c,d,e)ba, e(a,b)(c,d), c(b,e)(a,d)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, a(b,c)d, dabc, (c,d)ba, (a,b)(c,d), cb(a,d). In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d), a(b,c)d, cb(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, dabc, (c,d)ba, (a,b)(c,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e, ab(c,e), (c,e)ba, e(a,b)c, c(b,e)a. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab, ab(c,e), (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e, c(b,e)a.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d, (b,e)(a,d). In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d, (b,e)(a,d).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, cdea, (a,c,e)d, ac(d,e), da(c,e), (c,d,e)a, ea(c,d), ce(a,d). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, (a,c,e)d, da(c,e). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ac(d,e), (c,d,e)a, ea(c,d), ce(a,d).
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd, cdbe, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b, eb(c,d), c(b,e)d. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd, cdbe, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b, c(b,e)d.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, a(b,c)(d,e), dab(c,e), (c,d,e)ba, e(a,b)(c,d), c(b,e)(a,d)}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, dabc, (c,d)ba, (a,b)(c,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e, c(b,e)a, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d, (b,e)(a,d), eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ac(d,e), (c,d,e)a, ea(c,d), ce(a,d), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b, c(b,e)d

The next such element is bdea.

Next Stage 4 Winner	Stage 5 Winning Set	Rating

	bdea		bdeac			1
			bdeca			1
			bdcea			1
			bcdea			2
			cbdea			1
			bde(a,c)			1
			bd(c,e)a			3
			b(c,d)ea			2
			(b,c)dea			1
Try c(b,e)(a,d).

Covering Element		Covered Elements	How many times (cumulative)?

	c(b,e)(a,d)			abcd				2
					acde				2
					bcde				2
					abdc				2
					eabc				2
					eabd				2
					bc(a,d)			2
					c(a,d)e			2								cb(a,d)			2
					b(a,d,e)			2
					c(a,d,e)			2
					(a,d)bc			2
					(a,d)eb			2
					(a,d)ec			2
					debc				2
					e(a,d)b			2
 					e(a,d)c			2
					(a,d)cb			2
					eacb				2
					a(b,c)d			2
					eacd				2
					ab(c,e)			2
					b(c,e)d			2
					abed				2
					ba(c,e)			2
					bade				2
					bd(c,e)			2
					da(c,e)			2
					(c,e)ab			2
					deab				2
					d(c,e)a			2
					d(c,e)b			1
					(c,e)bd			2
					(c,e)da			2
					(c,e)ba			2
					e(a,b)c			2
					bdea				1
					eb(a,d)			1
					e(a,c,d)			2
					eb(c,d)			2
					(a,b,d)e			1
					cdbe				2
					(c,d)ea			1
					(a,c,e)d			2

Check to see that, upon reducing winning set from stage 5 to stage 4, there are no elements that are not in the stage 4 winning matrix that are covered more than once. Potential winning set: 
	
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, c(a,d)be, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, a(b,c)(d,e), dab(c,e), (c,d,e)ba, e(a,b)(c,d), c(b,e)(a,d)}

(1) Blot out an e: We get abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, acdb, abcd, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, abdc, b(a,c,d), (a,d)cb, (a,b,c,d), bc(a,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, a(b,c)d, dabc, (c,d)ba, (a,b)(c,d), cb(a,d). In winning set are: abcd, abdc, bc(a,d), cb(a,d), (a,d)bc, (a,d)bc, (a,d)cb, a(b,c)d, abcd, abdc, (a,d)cb, bc(a,d), a(b,c)d, cb(a,d). Not in winning set are: acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, dabc, (c,d)ba, (a,b)(c,d).
(2) Blot out a d: We get abce, eabc, bcae, cb(a,e), aebc, eabc, eacb, ea(b,c), eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, (c,e)ab, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e, ab(c,e), (c,e)ba, e(a,b)c, c(b,e)a. In winning set are: eabc, eabc, eacb, eacb, ab(c,e), ba(c,e), ba(c,e), (c,e)ab, (c,e)ba, e(a,b)c, (c,e)ab, ab(c,e), (c,e)ba, e(a,b)c. Not in winning set are: abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e, c(b,e)a.
(3) Blot out a c: We get abde, eabd, b(a,d)e, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, eadb, abed, bade, bdae, deab, ebda, de(a,b), bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), a(b,d)e, (a,d)be, d(b,e)a, edab, deab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d, (b,e)(a,d). In winning set are: eabd, b(a,d,e), (a,d)eb, e(a,d)b, e(a,d)b, eabd, abed, bade, deab, bdea, bade, abed, eb(a,d), (a,d)eb, (a,b,d)e, b(a,d,e), deab. Not in winning set are: abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d, (b,e)(a,d).
(4) Blot out a b: We get acde, eadc, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, a(c,e)d, ad(c,e), da(c,e), d(c,e)a, (c,e)da, deac, deca, cade, aedc, e(a,c,d), (a,d)ec, (a,c,d)e, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, cdea, (a,c,e)d, ac(d,e), da(c,e), (c,d,e)a, ea(c,d), ce(a,d). In winning set are: acde, c(a,d)e, c(a,d,e), (a,d)ec, e(a,d)c, e(a,d)c, eacd, eacd, da(c,e), d(c,e)a, (c,e)da, e(a,c,d), (a,d)ec, c(a,d,e), acde, c(a,d)e, d(c,e)a, (c,e)da, (c,d)ea, e(a,c,d), (a,c,e)d, (a,c,e)d, da(c,e). Not in winning set are: eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ac(d,e), (c,d,e)a, ea(c,d), ce(a,d).
(5) Blot out an a: We get bcde, ebdc, bcde, cb(d,e), debc, edbc, edcb, e(b,c)d, ecdb, b(c,e)d, bd(c,e), bd(c,e), d(c,e)b, (c,e)bd, debc, bdec, cbde, bedc, eb(c,d), decb, (b,c,d)e, bc(d,e), c(b,d)e, cdbe, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, (c,e)bd, cdbe, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b, eb(c,d), c(b,e)d. In winning set are: bcde, bcde, debc, b(c,e)d, bd(c,e), bd(c,e), (c,e)bd, debc, eb(c,d), cdbe, d(c,e)b, (c,e)bd, cdbe, eb(c,d). Not in winning set are: ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b, c(b,e)d.

Winning set is now 
{abcde, eabdc, bc(a,d)e, cb(a,d,e), (a,d)ebc, e(a,d)bc, e(a,d)cb, ea(b,c)d, eacdb, ab(c,e)d, bad(c,e), bda(c,e), d(c,e)ab, (c,e)bda, de(a,b)c, bdeca, cbade, abedc, eb(a,c,d), (a,d)ecb, (a,b,c,d)e, bc(a,d,e), ac(b,d)e, da(c,e)b, d(b,c,e)a, (c,e)dab, (c,d)eab, e(a,c,d)b, (a,c,e)bd, cdbea, b(a,c,e)d, a(b,c)(d,e), dab(c,e), (c,d,e)ba, e(a,b)(c,d), c(b,e)(a,d)}

Elements not in stage 4 winning matrix, but covered only once: 
acdb, badc, bdac, dcab, cbda, d(a,b)c, bdca, cbad, b(a,c,d), (a,b,c,d), ac(b,d), c(a,d)b, d(b,c)a, edab, (c,d)ab, (a,c,d)b, (a,c)bd, cdba, b(a,c)d, dabc, (c,d)ba, (a,b)(c,d), abce, bcae, cb(a,e), aebc, ea(b,c), beca, cbae, abec, eb(a,c), aecb, (a,b,c)e, bc(a,e), acbe, cabe, (b,c,e)a, ceab, e(a,c)b, (a,c,e)b, cbea, b(a,c,e), a(b,c)e, c(b,e)a, abde, b(a,d)e, eadb, bdae, ebda, de(a,b), a(b,d)e, (a,d)be, d(b,e)a, edab, e(a,d)b, (a,e)bd, dbea, b(a,e)d, ab(d,e), dabe, (d,e)ba, e(a,b)d, (b,e)(a,d), eadc, a(c,e)d, ad(c,e), deac, deca, cade, aedc, (a,c,d)e, cdea, ac(d,e), (c,d,e)a, ea(c,d), ce(a,d), ebdc, cb(d,e), edbc, edcb, e(b,c)d, ecdb, bdec, cbde, bedc, decb, (b,c,d)e, bc(d,e), c(b,d)e, d(b,c,e), (c,e)db, (c,d)eb, e(c,d)b, b(c,e)d, (b,c)(d,e), db(c,e), (c,d,e)b, c(b,e)d