List of axioms:
$(A1)$ - All Tautologies.
$(A2)$ - $(\forall x(\alpha \rightarrow \beta) \rightarrow (\forall x \alpha \rightarrow \forall x \beta)) $.
$(A3)$ - $(\forall x \alpha) \rightarrow [\alpha]^{t}_{x} $ if x is free for t in $\alpha$.
$(A4)$ - $\alpha \rightarrow (\forall x \alpha) $ if x is not free in $\alpha $.
$(A5)$ - $ x = x$
$(A6)$ - $(x=y) \rightarrow (\alpha \rightarrow \beta)$ if $\beta$ if obtained by substitution of at least one occurrence of x free for y in $\alpha$.
Modus Ponens as inference rule.
And Generalization Theorem:
If $\Gamma \vdash \phi$ and x do not occur free in any formula in $\Gamma$, then $\Gamma \vdash \forall x \phi$
In order to prove:
$\vdash (\alpha \rightarrow \exists x \beta) \leftrightarrow \exists x (\alpha \rightarrow \beta)$
I am trying to prove (consider x does not occur free in $\alpha$):
$\vdash (\alpha \rightarrow \exists x \beta) \rightarrow \exists x (\alpha \rightarrow \beta)$
$\vdash \exists x (\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow \exists x \beta) $
All the resolutions I saw uses Rule T (page 118) Enderton in the middle of the proof in order to prove 1 and 2. Is it possible to do without it?
Suposing 1.
- $(\alpha \rightarrow \exists x \beta)$
- $\forall x ¬(\alpha \rightarrow \beta) \rightarrow ¬(\alpha \rightarrow \beta)$ (A3)
Then i am trying to use the tautology: $¬(\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow ¬\beta)$ but without success to reach $\exists x(\alpha \rightarrow \beta)$.
