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

I think it's probably that if you require ${\displaystyle F}$ 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