TwoRound Voting
This a voting mechanism from Voting Theory. It is essentially the same as Plurality Voting, but run in two rounds
Given set $A$ of candidates
 Round 1: Using plurality voting mechanism, choose two candidates $a, b \in A$
 Round 2: Plurality voting only between the two $a$ and $b$
Example
 here we assume that preferences are stable: people don't change their preferences between two rounds
 Round 1:
 $a > b > c$  11 votes
 $b > a > c$  8 votes
 $c > b > a$  2 votes
 $a$ and $b$ win the 1st round
 Round 2:
 (we just remove $c$ from the previous rankings)
 $a > b$  11 votes
 $b > a$  8 + 2 votes
Criteria
This method satisfies:
This method does not satisfy:
$N = 16$ and $A = \{x, y, z\}$
We have the following individual preferences:
 6 voters $x > y > z$
 5 voters $z > x > y$
 4 voters $y > z > x$
 2 voters $y > x > z$
Elections:
 Round 1: $x$ and $y$ win ($x=6, z=5, y=6$)
 Round 2: $x$ wins (6+5: $x > y$, 6: $y > x$)
But suppose that $x$ manages to also convince the last two voters that he is better:
 6 voters $x > y > z$
 5 voters $z > x > y$
 4 voters $y > z > x$
 2 voters $x > y > z$
Note that $x$ by improving his position should remain the winner
Elections:
 Round 1: $x$ and $z$ ($x=6+2, z=5, y=4$)
 Round 2: $z$ wins! (not $x$!) (6+2: $x > y$, 9: $z > x$)
This counterexample shows that the Monotonicity principle is not respected by TwoRound Voting method.
Suppose we run an election in Belgium
 there are 2 communes  2 regions
 we have 3 candidates: $A = \{a, b, c\}$
 $N = 13$ for each region

Region I 
Region II

preferences

 4: $a > b > c$
 3: $b > a > c$
 3: $c > a > b$
 3: $c > b > a$

 4: $a > b > c$
 3: $c > a > b$
 3: $b > a > c$
 3: $b > a > c$

Round 1

${\color{blue}{a: 4}}, b: 3, {\color{blue}{c: 6}}$

${\color{blue}{a: 4}}, {\color{blue}{b: 6}}, c: 3$

Round 2

${\color{blue}{a: 7}}, c: 6$

${\color{blue}{a: 7}}, b: 6$

In both regions $a$ wins.
But if we consider the global region, we'll have different results:
 Round 1: ${\color{red}{a: 8}}, {\color{blue}{b: 9, c: 9}}$  note that $a$ loses and doesn't go to the next round
 Round 2: ${\color{blue}{b: 17}}, c: 9$
 $c$ wins
So the separability principle is not satisfied in this example.
Is not satisfied
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