Talk:Identity of indiscernibles

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Hack to prove identity of indiscernibles edit

Latest comment: 7 months ago1 comment1 person in discussion

If you allow arbitrary predicates in the definition of indiscernibility, then identity of indiscernibles is easily proven. Given $x$ and $y$ such that $\forall F(Fx\leftrightarrow Fy)$ , one can instantiate $F$ as $\lambda z.x=z$ , and thus we have $x=x\leftrightarrow x=y$ , which proves that $x=y$ . I feel like Max Black missed something here, but I'm not sure what.

I think it's probably that if you require $F$ to have no free variables, this trick doesn't work, but I feel like "identity of things that are in the same orbit of an automorphism" is too obviously false. NoLongerBreathedIn (talk) 01:05, 16 September 2023 (UTC)Reply

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