I'm looking for a counterexample for the following (false) equality. It should exist.
$n$ is an odd integer.
$f(x_1, x_2 .. x_n)$ is a boolean expression in 3 CNF form with boolean variables x.
$A = \exists x_1 \forall x_2 \exists x_3 \forall x_4 ... \forall x_{n-1} \exists x_n: f(x_1..x_n)$
Note that the exists and forall are alternating and that the expression begins and ends with exist.
$B = \forall x_2 \forall x_4 \forall x_6 .. \forall x_{n-1} \exists x_1 \exists x_3 .. \exists x_n : f(x_1..x_n)$
Somehow I have a flawed proof that implies $A=B$. I'm looking for a simple counterexample, a 3 CNF formula with few variables, so I can find the flaw in my proof.