Can a valid FOL formula be falsifiable? Suppose I have a FOL formula that I can prove is valid. Does the definition of validity exclude the possibility that a formula is falsifiable?
Here's an example:
exists x, y. (p(x, y) -> (p(y,x) -> forall z. p(z,z)))

If I perform validity check, it turns out that the negation is not satisfiable because you derive the following:
forall x, y. (p(x,y) and (p(y,x) and exists z. not p(z,z)))

However, if I define p to be not equal, then I can see that p(z, z) should resolve to false. Then, there exists an x, y such that the whole formula is falsifiable, because not equal returns true for any unique pair of x and y, and false always for z. 
 A: No, it can't.
The negation of the original formula is:

$\forall x, y \ [p(x,y) \land p(y,x) \land \exists z \ \lnot p(z,z)]$.

You are suggesting an interpretation with $\ne$ for the binary predicate $p$ that satisfy this formula (which is unsatisfiable, being the negation of a valid formula).
If we replace $p$ with $\ne$ we get:

$\forall x, y \ [(x \ne y) \land (y \ne x) \land \exists z \ (z = z)]$.

But in no not-empty domain $D$ whatever we can satisfy it: consider $a \in D$ and instantiate the leading quantifiers with $a$ to get:

$(a \ne a) \land (a \ne a) \land \textbf{ exists z } (z = z)$.

Thus, it is not true that: "for every $a \in D, \ldots$".
But this holds for every $D$ (that is not-empty) and thus there is no domain where the suggested interpretation of $p$ will satisfy the above formula.
In your original paraphrasing, you missed that the negation in front of the formula $p(z,z)$, when used with $\ne$ in place of $p$, resolves to just equality, in which case $z = z$ is always true.
