The axiom of choice as a consequence of a stronger semantics?

First we show that $\in$ is not a semantic symbol. Suppose that $\in$ is a semantic symbol.

Contradiction: We know that $x\in \left \{ y \right \}$ iff $x=y$ (syntactic definition) is true and $x\in \left \{ y \right \}$ iff $x\Leftrightarrow y$ (semantic definition) is not true, for example: $\emptyset\in \left \{ 0 \right \}$ iff $\emptyset \Leftrightarrow 0$ is not true.

Note: $x=y$ implies $x\Leftrightarrow y$, but not always vice versa.

Until we select $x$ (syntactically) in the expression $\forall y\exists x(x\in y)$, it can be semantically true and $\in$ will be a semantic symbol, but as we showed above this is not true, so we can't do not choose $x$. If we choose $x$, then we can take two different singletons of $y$ and both of them can not simultaneously be contain $x$ without semantics.

I want to notice remove and close questions the same people: Andrés E. Caicedo, Todd Trimble.

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