How can one formulation of the axiom of extension define set equality and the second one not?

Asked Apr 12 '16 at 10:29

Active Apr 12 '16 at 10:29

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I've seen the claim that there are two ways of writing the axiom of extension. The first one is ${\forall}A{\forall}B({\forall}x(x{\in}A{\iff}x{\in}B)){\iff}A=B$. This one supposedly admits $=$ as a primitive concept. The second one is ${\forall}A{\forall}B{\forall}x(A=B{\land}A{\in}x){\implies}B{\in}x)$.

The second one supposedly defines $=$. How? What qualities does the second definition possess that make it able to define set equality?

asked Apr 12 '16 at 10:29

jjb

Unless I misunderstand, your question seems to imply that the first one can't be used to define set equality ; but clearly it can, and it's the definition that is generally used. – Captain Lama Apr 12 '16 at 10:34
See Axiom of extensionality for the correct formulation. – Mauro ALLEGRANZA Apr 12 '16 at 11:28
For set theory formalized in first-order language without equality, we can define $=$ as follows: $\forall x \ (x \in A \iff x \in B) \Leftrightarrow (A=B)$. This definition expands the language with the new symbol for equality (a binary predicate) and licenses the replacement of the formula in the LHS with the abbreviation: $A=B$. – Mauro ALLEGRANZA Apr 12 '16 at 11:32
Thus, being the first formula the definition of equality, it does not admits $=$ as a primitive concept. – Mauro ALLEGRANZA Apr 12 '16 at 13:25
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See this related question: Different axiomatizations of set equality. – Vej Kse Apr 12 '16 at 15:08

How can one formulation of the axiom of extension define set equality and the second one not?

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