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I know that ZF is not finitely axiomatizable, but what about Z (i.e. ZF without Replacement)?

1 Answers 1

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No. It is not.

You can find the proof as Theorem 8 in:

Mathias A. R. The Strength of Mac Lane Set Theory, Annals of Pure and Applied Logic, 110 (2001) 107--234.

(The article also appears on Mathias' homepage without the need for a paywall)

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    @Sumac - see Hao Wang, [A Survey of mathematical logic](https://books.google.it/books), Chapter XVII: Relative Strength and Reducibility, page 438.2017-05-10