Prove that if a formula has property , then the formula (∃ ) also has property .

Question

I know that this statement is true, but don't know how to approach the proof.

Answer 1

Two cases:

(i) $x \ne y$, and so $x$ is free in $\exists y \ \beta$.

But then $x$ is free in $\beta$, and thus we have to apply the fact that $\beta$ has property $P$.

(ii) $x=y$. Thus: $(\exists y \ \beta)[a/y]= \exists y \ \beta$.

This means that $(\exists y \ \beta)[a/y]^{(I,E)}= (\exists y \ \beta)^{(I,E)}$.

Clearly $(\exists y \ \beta)^{(I,E)}$ is true iff some "$x$-variant" of $E$ satisfy $\exists y \ \beta$ in $I$. But an "$x$-variant" of $E$ is also an "$x$-variant" of $E(x \leftarrow a^I)$.

Thus:

$(\exists y \ \beta)[a/y]^{(I,E)}=(\exists y \ \beta)^{(I,E(x \leftarrow a^I))}$