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**emakarov** Yes. The formula $\displaystyle \exists x( \neg A(x) \vee B(x))$ is equivalent to $\displaystyle (\exists x\,\neg A(x)) \vee (\exists x\,B(x))$, while $\displaystyle (\neg \exists x\,A(x)) \vee (\exists x\,B(x))$ is equivalent to $\displaystyle (\forall x\,\neg A(x))\lor(\exists x\,B(x))$. So $\displaystyle \alpha$ does not derive $\displaystyle \gamma$ because $\displaystyle \exists x\,\neg A(x)$ does not derive $\displaystyle \forall x\,\neg A(x)$. A countermodel should have $\displaystyle \neg A(x)$ true for some x but not for all x, which is what you have done.