$∀x∃y (P(x, y) ∧ y \neq c ) ∧ ∀x∀y∀z (x = y ∨ ¬P(x, z) ∨ ¬P(y, z))$
What's a predicate $P$ and constant $c$ that would show this is satisfiable on the naturals numbers. And what's the proof that it's not satisfiable on a finite, non-empty universe?
P1: $∀x(s(x) \neq 0)$
P2: $∀x∃y(s(x) = s(y) \implies x = y)$
P3: $∀x(x + 0 = x)$
P4: $∀x∃y(x + s(y) = s(x + y))$
What's the proof that $∀x∀y (x + y) = (y + x)$ is not a logical consequence of P1, P2, P3, P4?
Hint.
Let $c_0=0$, then there is $c_1\neq c_0$ s.t. $P(c_0,c_1)$. Inductively, if $c_n$ is given then there is $c_{n+1}$ s.t. $P(c_n,c_{n+1})$. Consider $\{c_n:n\in\Bbb{N}\}$. Use the sentence $$\forall x\forall y\forall z P(x,z)\to x=y \lor \lnot P(y,z)$$ to prove that $c_n\neq c_m$ for all $n>m$. (Fix $m$ and use the induction for $n$.)
Consider the set of all ordinals below $\omega^2$, and interpret $s$ and $+$ as ordinal successor and ordinal addition.