I have a question about a variation of the content in this answer: <https://math.stackexchange.com/a/694319/418401>

It states (variated) that: 
$$
(\forall x)(\forall y)(\forall z) [ \lnot A(x,x)\land (A(x,y)\land A(y,z) \rightarrow A(x,z) ) \land (A(x,y) \lor A(y,x))] \land (\forall y)(\exists x)\lnot A(y,x)
$$

This sentence must be false for all finite domains, but is true for some infinite domain (indeed, take the set of integers and define $A(y,x)$ as $y > x$, then this implication is true). 

However, how can I proof that the form is false for a domain with $n$ elements? Then, by the last part of the disjunction, for every $y$, there exist $0 < k \leq n$ elements $x\_k$ where $A(y,x\_k)$ is true. Also, there are $i= n-k$ elements $x\_i$ where $A(y,x\_i)$ is false. But this doesn't say anything about the part of the form that says $(A(x,y) \land A(y,x))$ except that $A(x,y)$ must be true for some $x,y$..

Note that this is the negation of the form
$$
(\forall x)(\forall y)(\forall z) [ \lnot A(x,x)\land (A(x,y)\land A(y,z) \rightarrow A(x,z) ) \land (A(x,y) \lor A(y,x))] \implies (\exists y)(\forall x)A(y,x)
$$
Maybe we must proof that this is always true, but then again I can't make the next step in my thinking.