Counterexample to show that a formula is not logically valid?

Question

How to find a counterexample to show that the wfs is not logically valid ?

This is the formula $$[\forall x\forall y(A(x,y)\to A(y,x))\land\forall x\forall y\forall z(A(x,y)\land A(y,z)\to A(x,z))]\to\forall x A(x,x)$$

I tried to construct the interpretation as this:

Domain=$\mathbb N, A(x,y)=x<y$ and with this I get this $\forall x A(x,x)$ false but $[\forall x\forall y(A(x,y)\to A(y,x))\land\forall x\forall y\forall z(A(x,y)\land A(y,z)\to A(x,z))]$ is false too, and it should be true.

How can I find such interpretation? Is there a trick on how to find it?

When we talk about just one variable like this $A(x)$ the common counterexamples are like $A(x)= "x $ is even", but I don't know what to do when it's whit 2 variables like this $A(x,y)$.

Answer 1

The formula claims that:

If a relation $A$ is symmetric and transitive, then it must be reflexive.

The three properties of being reflexive, symmetric, and transitive are well-known to be fully independent, and it's possible to construct relations that have any subset of the three.

For our counterexample, take the domain to be $\mathbb N$, and define $A(x, y) \equiv$ "$x$ and $y$ are both even". Then $A$ is symmetric and transitive, but not reflexive (e.g.: $A(1, 1)$ is false).