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What are the pros and cons of working with first-order logic with equality for constructing ZFC, when all you have to do is make '$x=y$' a shorthand for: $$'\forall z [z \in x \Leftrightarrow z \in y] \land \forall w [x \in w \Leftrightarrow y \in w]'$$

Constantine
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    I feel like this is not quite a replacement for logical equality -- in "ordinary" ZFC you would like to be able to conclude $\forall w [x \in w \Leftrightarrow y \in w]$ from $\forall z [z \in x \Leftrightarrow z \in y]$. It seems that you are not able to do this from this. (I think, instead, you'd want to define equality as only the first of the conjuncts, and then add an axiom stating that the first conjunct implies the second.) – Mees de Vries Apr 14 '16 at 09:21
  • How else do you determine if two sets are equal (the same) if not by looking at their respective elements? – Dan Christensen Apr 14 '16 at 13:29

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There are two "flavours" of first-order logic: with and without equality.

In the first case (with), the equality symbol is considered a logical symbol, i.e. we cannot "interpret" it in different ways according to the context (like the conncetive and the quantifiers).

In the second one (without) the equality symbol is considered a "mathematical" symbol, like $+$ for f-o arithmetic, and thus (in principle) may be interpreted in different ways according to the context.

If we chose to develop set theory in f-o language without equality, we have to expand the "basic" language with the new (binary) predicate symbol $=$ defined by:

$(x=y) ↔ ∀z(z∈x ↔ z∈y)$.

Mauro ALLEGRANZA
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  • So, you're saying that we use FOL with equality to prohibit people from giving interpretations to '=', at least in the case of ZFC? – Constantine Apr 14 '16 at 09:52
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    @Constantine - more generally, we use FOL with equality because, for the purpose of formalising mathematics, it’s natural to interpret the symbol $=$ by actual equality on the domain of the structure we are using to interpret the language. – Mauro ALLEGRANZA Apr 14 '16 at 10:06
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    Is your definition of "$=$" enough? Or do you also need "$\land \forall z ( x \in z \leftrightarrow y \in z )$"? – user21820 Aug 07 '18 at 14:22