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This is taken from Open Logic Project, p. 214, Release 2020-06-25.

Extensionality is formulated as:

$$ \forall x \forall y ( \forall z (z \in x \leftrightarrow z \in y) \to x=y). $$

Now, by the definition of implication ($\to$), $\forall z (z \in x \leftrightarrow z \in y)$ can be false while $x = y$ is true and the implication is true. But this is absurd, as that means that the statement: "Two sets which do not have all their elements in common are the same set," is valid, though it is clearly not. I must be missing something big here. Where is my reasoning faulty?

Alex Kruckman
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God bless
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  • I'm not confident that the answers to the linked question will resolve this asker's confusion. It would probably be helpful to explain why it's false that "by the definition of implication ($\to$), $\forall z (z \in x \leftrightarrow z \in y)$ can be false while $x = y$ is true". – Tanner Swett Jul 29 '20 at 11:56
  • @TannerSwett No need, I understand now, thank you nonetheless. The biconditional is not necessary due to how variable assignment is defined: if the term $x = y$ is true for some variable assignment $s$ in a structure $\mathfrak{M}$, then $s$ must assign the same value to $x$ and $y$, hence $\forall z (z \in x \leftrightarrow z \in y)$ is necessarily also true. – God bless Jul 29 '20 at 15:43

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