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I think it doesn't, because if $\Gamma \vDash \alpha \iff \Gamma \vDash \beta$,

then, by definition: $v(\Gamma) = 1 \to v(\alpha) = 1 \iff v(\Gamma) = 1 \to v(\beta) = 1$,

which is the same as: $(v(\Gamma) = 1 \to v(\alpha) = 1) \land v(\Gamma) = 1 \iff v(\beta) = 1$,

and we can simplify as: $(v(\Gamma) = 1 \land v(\alpha) = 1) \iff v(\beta) = 1$.

And that looks different than $v(\alpha) = 1 \iff v(\beta) = 1$.

Actually, for the case that $(v(\Gamma) = 0 \land v(\alpha) = 1)$, the assumption demands that $v(\beta) = 0$, right?

In this case, $\alpha \iff \beta$ doesn't hold, since they have different values...but I'm not sure about this conclusion though.

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No it doesn't. Example: whenever p is true then p or q is true. Whenever p is true p is true. But it is not true that p iff p or q.

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