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

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?)

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    $\begingroup$ 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. $\endgroup$ – hardmath Apr 13 at 5:15
  • $\begingroup$ 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. $\endgroup$ – Bram28 Apr 21 at 16:51
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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 $0 \ngtr \bigstar_0$.

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