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Give an equational proof $$ \vdash (\forall x)(A \rightarrow (\exists x)B) \equiv ((\exists x)A \rightarrow (\exists x)B)$$

I don't know where to start. Maybe I could start with $ (\forall x)(A \rightarrow (\exists x)B) $ and change it to $ (\forall x)(\lnot A \lor (\exists x)B)$.

See George Tourlakis, Mathematical Logic (2008) or this post for a list of axioms and theorems.

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Out Of Bounds
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1 Answers1

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As you have proposed, transform : $(∀x)(A \rightarrow (∃x)B)$ into :

$(∀x)(\lnot A \lor (∃x)B)$.

Then, apply 6.4.2 Corollary : $\vdash (\forall x)(A \lor B) \equiv A \lor (\forall x)B$, provided $x$ is not free in $A$,

where $x$ is not free in $(∃x)B$, to get :

$(∀x)\lnot A \lor (∃x)B$.

Then we "switch" to $\exists$ :

$\lnot(\exists x)A \lor (∃x)B$.

Finally, we reintroduce $\rightarrow$ :

$(\exists x)A \rightarrow (∃x)B$.

All steps are equivalences.

Mauro ALLEGRANZA
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