# Prove trichotomy of order WITHOUT induction?

Claim. If $$x,y \in \mathbb N$$, then at least one of these is true: (a) $$x>y$$; (b) $$x=y$$; or (c) $$x.

It seems that the usual proof of this claim uses induction. Is it possible to prove this without induction? (Or is perhaps this claim false without the axiom of induction?)

• Your Question would be more meaningful if it spelled out what assumptions are left after omitting induction, and how the relations $\lt,=,\gt$ are accordingly defined. – hardmath Apr 13 at 5:15
• You really need to provide the formal definitions of $<$ and $>$ in order to get a proper answer to this. Depending on that, counterexamples may or may not work. – Bram28 Apr 21 at 16:51

If we drop the induction axiom (but keep the other four Peano axioms), then we could have $$\mathbb N = \{0,\bigstar_0,1,\bigstar_1,2,\bigstar_2,\dots \}.$$
And now $$0 \nless \bigstar_0$$ , $$0\neq \bigstar_0$$, and $$0 \ngtr \bigstar_0$$.