A voting system is consistent if, whenever the electorate is divided (arbitrarily) into several parts and elections in those parts garner the same result, then an election of the entire electorate also garners that result. Smith^{[1]} calls this property separability and Woodall^{[2]} calls it convexity.
It has been proven a ranked voting system is "consistent if and only if it is a scoring function"^{[3]}, i.e. a positional voting system. Borda count is an example of this.
The failure of the consistency criterion can be seen as an example of Simpson's paradox.
As shown below under KemenyYoung, passing or failing the consistency criterion can depend on whether the election selects a single winner or a full ranking of the candidates (sometimes referred to as ranking consistency); in fact, the specific examples below rely on finding single winner inconsistency by choosing two different rankings with the same overall winner, which means they do not apply to ranking consistency.
Examples
Copeland
This example shows that Copeland's method violates the consistency criterion. Assume five candidates A, B, C, D and E with 27 voters with the following preferences:
Preferences  Voters 

A > D > B > E > C  3 
A > D > E > C > B  2 
B > A > C > D > E  3 
C > D > B > E > A  3 
E > C > B > A > D  3 
A > D > C > E > B  3 
A > D > E > B > C  1 
B > D > C > E > A  3 
C > A > B > D > E  3 
E > B > C > A > D  3 
Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.
First group of voters
In the following the Copeland winner for the first group of voters is determined.
Preferences  Voters 

A > D > B > E > C  3 
A > D > E > C > B  2 
B > A > C > D > E  3 
C > D > B > E > A  3 
E > C > B > A > D  3 
The results would be tabulated as follows:
X  
A  B  C  D  E  
Y  A  [X] 9 [Y] 5 
[X] 6 [Y] 8 
[X] 3 [Y] 11 
[X] 6 [Y] 8  
B  [X] 5 [Y] 9 
[X] 8 [Y] 6 
[X] 8 [Y] 6 
[X] 5 [Y] 9  
C  [X] 8 [Y] 6 
[X] 6 [Y] 8 
[X] 5 [Y] 9 
[X] 8 [Y] 6  
D  [X] 11 [Y] 3 
[X] 6 [Y] 8 
[X] 9 [Y] 5 
[X] 3 [Y] 11  
E  [X] 8 [Y] 6 
[X] 9 [Y] 5 
[X] 6 [Y] 8 
[X] 11 [Y] 3 

Pairwise election results (wontiedlost):  301  202  202  202  103 
 [X] indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
 [Y] indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption
Result: With the votes of the first group of voters, A can defeat three of the four opponents, whereas no other candidate wins against more than two opponents. Thus, A is elected Copeland winner by the first group of voters.
Second group of voters
Now, the Copeland winner for the second group of voters is determined.
Preferences  Voters 

A > D > C > E > B  3 
A > D > E > B > C  1 
B > D > C > E > A  3 
C > A > B > D > E  3 
E > B > C > A > D  3 
The results would be tabulated as follows:
X  
A  B  C  D  E  
Y  A  [X] 6 [Y] 7 
[X] 9 [Y] 4 
[X] 3 [Y] 10 
[X] 6 [Y] 7  
B  [X] 7 [Y] 6 
[X] 6 [Y] 7 
[X] 4 [Y] 9 
[X] 7 [Y] 6  
C  [X] 4 [Y] 9 
[X] 7 [Y] 6 
[X] 7 [Y] 6 
[X] 4 [Y] 9  
D  [X] 10 [Y] 3 
[X] 9 [Y] 4 
[X] 6 [Y] 7 
[X] 3 [Y] 10  
E  [X] 7 [Y] 6 
[X] 6 [Y] 7 
[X] 9 [Y] 4 
[X] 10 [Y] 3 

Pairwise election results (wontiedlost):  301  202  202  202  103 
Result: Taking only the votes of the second group in account, again, A can defeat three of the four opponents, whereas no other candidate wins against more than two opponents. Thus, A is elected Copeland winner by the second group of voters.
All voters
Finally, the Copeland winner of the complete set of voters is determined.
Preferences  Voters 

A > D > B > E > C  3 
A > D > C > E > B  3 
A > D > E > B > C  1 
A > D > E > C > B  2 
B > A > C > D > E  3 
B > D > C > E > A  3 
C > A > B > D > E  3 
C > D > B > E > A  3 
E > B > C > A > D  3 
E > C > B > A > D  3 
The results would be tabulated as follows:
X  
A  B  C  D  E  
Y  A  [X] 15 [Y] 12 
[X] 15 [Y] 12 
[X] 6 [Y] 21 
[X] 12 [Y] 15  
B  [X] 12 [Y] 15 
[X] 14 [Y] 13 
[X] 12 [Y] 15 
[X] 12 [Y] 15  
C  [X] 12 [Y] 15 
[X] 13 [Y] 14 
[X] 12 [Y] 15 
[X] 12 [Y] 15  
D  [X] 21 [Y] 6 
[X] 15 [Y] 12 
[X] 15 [Y] 12 
[X] 6 [Y] 21  
E  [X] 15 [Y] 12 
[X] 15 [Y] 12 
[X] 15 [Y] 12 
[X] 21 [Y] 6 

Pairwise election results (wontiedlost):  202  301  400  103  004 
Result: C is the Condorcet winner, thus Copeland chooses C as winner.
Conclusion
A is the Copeland winner within the first group of voters and also within the second group of voters. However, both groups combined elect C as the Copeland winner. Thus, Copeland fails the consistency criterion.
Instantrunoff voting
This example shows that Instantrunoff voting violates the consistency criterion. Assume three candidates A, B and C and 23 voters with the following preferences:
Preferences  Voters 

A > B > C  4 
B > A > C  2 
C > B > A  4 
A > B > C  4 
B > A > C  6 
C > A > B  3 
Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.
First group of voters
In the following the instantrunoff winner for the first group of voters is determined.
Preferences  Voters 

A > B > C  4 
B > A > C  2 
C > B > A  4 
B has only 2 votes and is eliminated first. Its votes are transferred to A. Now, A has 6 votes and wins against C with 4 votes.
Candidate  Votes in round  

1st  2nd  
A  4  6 
B  2  
C  4  4 
Result: A wins against C, after B has been eliminated.
Second group of voters
Now, the instantrunoff winner for the second group of voters is determined.
Preferences  Voters 

A > B > C  4 
B > A > C  6 
C > A > B  3 
C has the fewest votes, a count of 3, and is eliminated. A benefits from that, gathering all the votes from C. Now, with 7 votes A wins against B with 6 votes.
Candidate  Votes in round  

1st  2nd  
A  4  7 
B  6  6 
C  3 
Result: A wins against B, after C has been eliminated.
All voters
Finally, the instant runoff winner of the complete set of voters is determined.
Preferences  Voters 

A > B > C  8 
B > A > C  8 
C > A > B  3 
C > B > A  4 
C has the fewest first preferences and so is eliminated first, its votes are split: 4 are transferred to B and 3 to A. Thus, B wins with 12 votes against 11 votes of A.
Candidate  Votes in round  

1st  2nd  
A  8  11 
B  8  12 
C  7 
Result: B wins against A, after C is eliminated.
Conclusion
A is the instantrunoff winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the instantrunoff winner. Thus, instantrunoff voting fails the consistency criterion.
KemenyYoung method
This example shows that the Kemeny–Young method violates the consistency criterion. Assume three candidates A, B and C and 38 voters with the following preferences:
Group  Preferences  Voters 

1st  A > B > C  7 
B > C > A  6  
C > A > B  3  
2nd  A > C > B  8 
B > A > C  7  
C > B > A  7 
Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.
First group of voters
In the following the KemenyYoung winner for the first group of voters is determined.
Preferences  Voters 

A > B > C  7 
B > C > A  6 
C > A > B  3 
The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:
Pairs of choices  Voters who prefer  

X  Y  X over Y  Neither  Y over X 
A  B  10  0  6 
A  C  7  0  9 
B  C  13  0  3 
The ranking scores of all possible rankings are:
Preferences  1 vs 2  1 vs 3  2 vs 3  Total 

A > B > C  10  7  13  30 
A > C > B  7  10  3  20 
B > A > C  6  13  7  26 
B > C > A  13  6  9  28 
C > A > B  9  3  10  22 
C > B > A  3  9  6  18 
Result: The ranking A > B > C has the highest ranking score. Thus, A wins ahead of B and C.
Second group of voters
Now, the KemenyYoung winner for the second group of voters is determined.
Preferences  Voters 

A > C > B  8 
B > A > C  7 
C > B > A  7 
The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:
Pairs of choices  Voters who prefer  

X  Y  X over Y  Neither  Y over X 
A  B  8  0  14 
A  C  15  0  7 
B  C  7  0  15 
The ranking scores of all possible rankings are:
Preferences  1 vs 2  1 vs 3  2 vs 3  Total 

A > B > C  8  15  7  30 
A > C > B  15  8  15  38 
B > A > C  14  7  15  36 
B > C > A  7  14  7  28 
C > A > B  7  15  8  30 
C > B > A  15  7  14  36 
Result: The ranking A > C > B has the highest ranking score. Hence, A wins ahead of C and B.
All voters
Finally, the KemenyYoung winner of the complete set of voters is determined.
Preferences  Voters 

A > B > C  7 
A > C > B  8 
B > A > C  7 
B > C > A  6 
C > A > B  3 
C > B > A  7 
The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:
Pairs of choices  Voters who prefer  

X  Y  X over Y  Neither  Y over X 
A  B  18  0  20 
A  C  22  0  16 
B  C  20  0  18 
The ranking scores of all possible rankings are:
Preferences  1 vs 2  1 vs 3  2 vs 3  Total 

A > B > C  18  22  20  60 
A > C > B  22  18  18  58 
B > A > C  20  20  22  62 
B > C > A  20  20  16  56 
C > A > B  16  18  18  52 
C > B > A  18  16  20  54 
Result: The ranking B > A > C has the highest ranking score. So, B wins ahead of A and C.
Conclusion
A is the KemenyYoung winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the KemenyYoung winner. Thus, the Kemeny–Young method fails the consistency criterion.
Ranking consistency
The KemenyYoung method satisfies ranking consistency; that is, if the electorate is divided arbitrarily into two parts and separate elections in each part result in the same ranking being selected, an election of the entire electorate also selects that ranking.
Informal proof
The KemenyYoung score of a ranking is computed by summing up the number of pairwise comparisons on each ballot that match the ranking . Thus, the KemenyYoung score for an electorate can be computed by separating the electorate into disjoint subsets (with ), computing the KemenyYoung scores for these subsets and adding it up:
 .
Now, consider an election with electorate . The premise of the consistency criterion is to divide the electorate arbitrarily into two parts , and in each part the same ranking is selected. This means, that the KemenyYoung score for the ranking in each electorate is bigger than for every other ranking :
Now, it has to be shown, that the KemenyYoung score of the ranking in the entire electorate is bigger than the KemenyYoung score of every other ranking :
Thus, the KemenyYoung method is consistent with respect to complete rankings.
Majority Judgment
This example shows that majority judgment violates the consistency criterion. Assume two candidates A and B and 10 voters with the following ratings:
Candidate  Voters  

A  B  
Excellent  Fair  3 
Poor  Fair  2 
Fair  Poor  3 
Poor  Fair  2 
Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.
First group of voters
In the following the majority judgment winner for the first group of voters is determined.
Candidates  Voters  

A  B  
Excellent  Fair  3 
Poor  Fair  2 
The sorted ratings would be as follows:
Candidate 
 
A 
 
B 
 
Excellent Good Fair Poor 
Result: With the votes of the first group of voters, A has the median rating of "Excellent" and B has the median rating of "Fair". Thus, A is elected majority judgment winner by the first group of voters.
Second group of voters
Now, the majority judgment winner for the second group of voters is determined.
Candidates  Voters  

A  B  
Fair  Poor  3 
Poor  Fair  2 
The sorted ratings would be as follows:
Candidate 
 
A 
 
B 
 
Excellent Good Fair Poor 
Result: Taking only the votes of the second group in account, A has the median rating of "Fair" and B the median rating of "Poor". Thus, A is elected majority judgment winner by the second group of voters.
All voters
Finally, the majority judgment winner of the complete set of voters is determined.
Candidates  Voters  

A  B  
Excellent  Fair  3 
Fair  Poor  3 
Poor  Fair  4 
The sorted ratings would be as follows:
Candidate 
 
A 
 
B 
 
Excellent Good Fair Poor 
The median ratings for A and B are both "Fair". Since there is a tie, "Fair" ratings are removed from both, until their medians become different. After removing 20% "Fair" ratings from the votes of each, the sorted ratings are now:
Candidate 
 
A 
 
B 

Result: Now, the median rating of A is "Poor" and the median rating of B is "Fair". Thus, B is elected majority judgment winner.
Conclusion
A is the majority judgment winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the Majority Judgment winner. Thus, Majority Judgment fails the consistency criterion.
Minimax
This example shows that the minimax method violates the consistency criterion. Assume four candidates A, B, C and D with 43 voters with the following preferences:
Preferences  Voters 

A > B > C > D  1 
A > D > B > C  6 
B > C > D > A  5 
C > D > B > A  6 
A > B > D > C  8 
A > D > C > B  2 
C > B > D > A  9 
D > C > B > A  6 
Since all preferences are strict rankings (no equals are present), all three minimax methods (winning votes, margins and pairwise opposite) elect the same winners.
Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.
First group of voters
In the following the minimax winner for the first group of voters is determined.
Preferences  Voters 

A > B > C > D  1 
A > D > B > C  6 
B > C > D > A  5 
C > D > B > A  6 
The results would be tabulated as follows:
X  
A  B  C  D  
Y  A  [X] 11 [Y] 7 
[X] 11 [Y] 7 
[X] 11 [Y] 7  
B  [X] 7 [Y] 11 
[X] 6 [Y] 12 
[X] 12 [Y] 6  
C  [X] 7 [Y] 11 
[X] 12 [Y] 6 
[X] 6 [Y] 12  
D  [X] 7 [Y] 11 
[X] 6 [Y] 12 
[X] 12 [Y] 6 

Pairwise election results (wontiedlost)  003  201  201  201  
Worst pairwise  Defeat (winning votes)  11  12  12  12 
Defeat (margins)  4  6  6  6  
Opposition  11  12  12  12 
 [X] indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
 [Y] indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption
Result: The candidates B, C and D form a cycle with clear defeats. A benefits from that since it loses relatively closely against all three and therefore A's biggest defeat is the closest of all candidates. Thus, A is elected minimax winner by the first group of voters.
Second group of voters
Now, the minimax winner for the second group of voters is determined.
Preferences  Voters 

A > B > D > C  8 
A > D > C > B  2 
C > B > D > A  9 
D > C > B > A  6 
The results would be tabulated as follows:
X  
A  B  C  D  
Y  A  [X] 15 [Y] 10 
[X] 15 [Y] 10 
[X] 15 [Y] 10  
B  [X] 10 [Y] 15 
[X] 17 [Y] 8 
[X] 8 [Y] 17  
C  [X] 10 [Y] 15 
[X] 8 [Y] 17 
[X] 16 [Y] 9  
D  [X] 10 [Y] 15 
[X] 17 [Y] 8 
[X] 9 [Y] 16 

Pairwise election results (wontiedlost)  003  201  201  201  
Worst pairwise  Defeat (winning votes)  15  17  16  17 
Defeat (margins)  5  9  7  9  
Opposition  15  17  16  17 
Result: Taking only the votes of the second group in account, again, B, C and D form a cycle with clear defeats and A benefits from that because of its relatively close losses against all three and therefore A's biggest defeat is the closest of all candidates. Thus, A is elected minimax winner by the second group of voters.
All voters
Finally, the minimax winner of the complete set of voters is determined.
Preferences  Voters 

A > B > C > D  1 
A > B > D > C  8 
A > D > B > C  6 
A > D > C > B  2 
B > C > D > A  5 
C > B > D > A  9 
C > D > B > A  6 
D > C > B > A  6 
The results would be tabulated as follows:
X  
A  B  C  D  
Y  A  [X] 26 [Y] 17 
[X] 26 [Y] 17 
[X] 26 [Y] 17  
B  [X] 17 [Y] 26 
[X] 23 [Y] 20 
[X] 20 [Y] 23  
C  [X] 17 [Y] 26 
[X] 20 [Y] 23 
[X] 22 [Y] 21  
D  [X] 17 [Y] 26 
[X] 23 [Y] 20 
[X] 21 [Y] 22 

Pairwise election results (wontiedlost)  003  201  201  201  
Worst pairwise  Defeat (winning votes)  26  23  22  23 
Defeat (margins)  9  3  1  3  
Opposition  26  23  22  23 
Result: Again, B, C and D form a cycle. But now, their mutual defeats are very close. Therefore, the defeats A suffers from all three are relatively clear. With a small advantage over B and D, C is elected minimax winner.
Conclusion
A is the minimax winner within the first group of voters and also within the second group of voters. However, both groups combined elect C as the Minimax winner. Thus, Minimax fails the consistency criterion.
Ranked pairs
This example shows that the Ranked pairs method violates the consistency criterion. Assume three candidates A, B and C with 39 voters with the following preferences:
Preferences  Voters 

A > B > C  7 
B > C > A  6 
C > A > B  3 
A > C > B  9 
B > A > C  8 
C > B > A  6 
Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.
First group of voters
In the following the Ranked pairs winner for the first group of voters is determined.
Preferences  Voters 

A > B > C  7 
B > C > A  6 
C > A > B  3 
The results would be tabulated as follows:
X  
A  B  C  
Y  A  [X] 6 [Y] 10 
[X] 9 [Y] 7  
B  [X] 10 [Y] 6 
[X] 3 [Y] 13  
C  [X] 7 [Y] 9 
[X] 13 [Y] 3 

Pairwise election results (wontiedlost):  101  101  101 
 [X] indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
 [Y] indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption
The sorted list of victories would be:
Pair  Winner 

B (13) vs C (3)  B 13 
A (10) vs B (6)  A 10 
A (7) vs C (9)  C 9 
Result: B > C and A > B are locked in first (and C > A can't be locked in after that), so the full ranking is A > B > C. Thus, A is elected Ranked pairs winner by the first group of voters.
Second group of voters
Now, the Ranked pairs winner for the second group of voters is determined.
Preferences  Voters 

A > C > B  9 
B > A > C  8 
C > B > A  6 
The results would be tabulated as follows:
X  
A  B  C  
Y  A  [X] 14 [Y] 9 
[X] 6 [Y] 17  
B  [X] 9 [Y] 14 
[X] 15 [Y] 8  
C  [X] 17 [Y] 6 
[X] 8 [Y] 15 

Pairwise election results (wontiedlost):  101  101  101 
The sorted list of victories would be:
Pair  Winner 

A (17) vs C (6)  A 17 
B (8) vs C (15)  C 15 
A (9) vs B (14)  B 14 
Result: Taking only the votes of the second group in account, A > C and C > B are locked in first (and B > A can't be locked in after that), so the full ranking is A > C > B. Thus, A is elected Ranked pairs winner by the second group of voters.
All voters
Finally, the Ranked pairs winner of the complete set of voters is determined.
Preferences  Voters 

A > B > C  7 
A > C > B  9 
B > A > C  8 
B > C > A  6 
C > A > B  3 
C > B > A  6 
The results would be tabulated as follows:
X  
A  B  C  
Y  A  [X] 20 [Y] 19 
[X] 15 [Y] 24  
B  [X] 19 [Y] 20 
[X] 18 [Y] 21  
C  [X] 24 [Y] 15 
[X] 21 [Y] 18 

Pairwise election results (wontiedlost):  101  200  002 
The sorted list of victories would be:
Pair  Winner 

A (25) vs C (15)  A 24 
B (21) vs C (18)  B 21 
A (19) vs B (20)  B 20 
Result: Now, all three pairs (A > C, B > C and B > A) can be locked in without a cycle. The full ranking is B > A > C. Thus, Ranked pairs chooses B as winner, which is the Condorcet winner, due to the lack of a cycle.
Conclusion
A is the Ranked pairs winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the Ranked pairs winner. Thus, the Ranked pairs method fails the consistency criterion.
Schulze method
This example shows that the Schulze method violates the consistency criterion. Again, assume three candidates A, B and C with 39 voters with the following preferences:
Preferences  Voters 

A > B > C  7 
B > C > A  6 
C > A > B  3 
A > C > B  9 
B > A > C  8 
C > B > A  6 
Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.
First group of voters
In the following the Schulze winner for the first group of voters is determined.
Preferences  Voters 

A > B > C  7 
B > C > A  6 
C > A > B  3 
The pairwise preferences would be tabulated as follows:
d[X, Y]  Y  

A  B  C  
X  A  10  7  
B  6  13  
C  9  3 
Now, the strongest paths have to be identified, e.g. the path A > B > C is stronger than the direct path A > C (which is nullified, since it is a loss for A).
d[X, Y]  Y  

A  B  C  
X  A  10  10  
B  9  13  
C  9  9 
Result: A > B, A > C and B > C prevail, so the full ranking is A > B > C. Thus, A is elected Schulze winner by the first group of voters.
Second group of voters
Now, the Schulze winner for the second group of voters is determined.
Preferences  Voters 

A > C > B  9 
B > A > C  8 
C > B > A  6 
The pairwise preferences would be tabulated as follows:
d[X, Y]  Y  

A  B  C  
X  A  9  17  
B  14  8  
C  6  15 
Now, the strongest paths have to be identified, e.g. the path A > C > B is stronger than the direct path A > B.
d[X, Y]  Y  

A  B  C  
X  A  15  17  
B  14  14  
C  14  15 
Result: A > B, A > C and C > B prevail, so the full ranking is A > C > B. Thus, A is elected Schulze winner by the second group of voters.
All voters
Finally, the Schulze winner of the complete set of voters is determined.
Preferences  Voters 

A > B > C  7 
A > C > B  9 
B > A > C  8 
B > C > A  6 
C > A > B  3 
C > B > A  6 
The pairwise preferences would be tabulated as follows:
d[X, Y]  Y  

A  B  C  
X  A  19  24  
B  20  21  
C  15  18 
Now, the strongest paths have to be identified:
d[X, Y]  Y  

A  B  C  
X  A  0  24  
B  20  21  
C  0  0 
Result: A > C, B > A and B > C prevail, so the full ranking is B > A > C. Thus, Schulze chooses B as winner. In fact, B is also Condorcet winner.
Conclusion
A is the Schulze winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the Schulze winner. Thus, the Schulze method fails the consistency criterion.
STAR voting
References
 ^ John H Smith, "Aggregation of preferences with variable electorate", Econometrica, Vol. 41 (1973), pp. 1027–1041.
 ^ D. R. Woodall, "Properties of preferential election rules", Voting matters, Issue 3 (December 1994), pp. 8–15.
 ^ H. P. Young, "Social Choice Scoring Functions", SIAM Journal on Applied Mathematics Vol. 28, No. 4 (1975), pp. 824–838.