I'm given the formula: ∃x∃y∃z(P(x, y) ∧ P(z, y) ∧ P(x, z) ∧ ¬P(z, x)).
The universe is all natural numbers. R is the relation corresponding to P. R = {<x, x+1> : x >= 0}
When putting values into the formula I've thought that if x = x then y = x + 1. By using this logic z is equal to x for P(z, y), but z has to be equal to y for P(x, z). I can't think of any other way to do this.
How do I show whether the formula is true or false? Could anyone point me in the right direction?
Indeed. That is the way to do this.
Let's tidy up your work a bit.
You have $P:=\{\langle x, x+1\rangle: x\in\Bbb N^+\}$ Which means $P(x,y)$ is substitutable with $y=x+1$. Okay, so we shall do that.
$$\exists x~\exists y~\exists z~(P(x, y) \land P(z, y) \land P(x, z) \land\lnot P(z, x))\\\equiv\\\exists x~\exists y~\exists z~(y=x+1 \land y=z+1 \land z=x+1 \land x\neq z+1)$$
Now, assuming this was true, we could find naturals $x,y,z$ where:
So...