Counterexample to see that a formula is $F$

Question

How to find an interpretation $I$ such that make the following formula false?

$$\forall x\forall y\forall z(A(x,x)\land (A(x,z)\to A(x,y)\lor A(y,z)))\to\exists y\forall z A(y,z)$$

I tried with 3 interpretations but all failed :(

$I_1$ with domain $\mathbb N, A(x,y)="x=y"$

$I_2$ with domain $\mathbb N, A(x,y)="x,y $ are both even"

$I_3$ with domain $\mathbb N, A(x,y)="x<y"$

Any help here is appreciated.

Answer 1

You're close! Try $A(x,y): x \ge y$