How can I prove this in the predicate logic?
$B \rightarrow \exists xA $ Eq $\exists x (B \rightarrow A)$
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It depends on the details of the proof system. For an Hilbert's style proof, see the answer to this post. – Mauro ALLEGRANZA Oct 16 '15 at 12:02
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Are you given that $x$ isn't free in $B$? If not, then the claimed result is wrong. – Andreas Blass Oct 16 '15 at 12:08
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Demonstrate $B\to \exists x\; A \;\vdash\; \exists x\;(B\to A)$, by natural deduction. $$\begin{array}{ll}B\to \exists x\;A & \text{P1} \\ \begin{array}{|ll} B & \text{Assume} \\\hline \exists x\;A & \text{Implication Elimination} \\ {A\mid}^x_c & \text{Existential Elimination} \end{array}\\ B\to {A\mid}^x_c & \text{Implication Introduction} \\ \exists x\;( B\to A ) & \text{Existential Introduction} \\ \Box\end{array}$$
Left to do: demonstrate $\exists x\;(B\to A) \;\vdash\; (B\to \exists x\;A)$
Graham Kemp
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