How to prove UG is sound? I want to show  that the PL is sound for the set of rules    $ S=\{P,T,C,US,UG,E \} $
That is, if $\Gamma \vdash_s \phi$, then $\Gamma \vDash \phi$
And I have already proved it except for UG
If $ \phi$ occurred in k+1 by UG, then there is $\phi (\alpha/\beta)$ in 1 < $ \mathcal l$  < k.
And by induction hypothesis, $\Gamma \vDash \phi (\alpha/\beta)$
My textbook, Benson mates, remains for me to show that for any  fomula $\phi $ and variable $\alpha$, if $\Gamma \vDash\phi(\alpha/\beta)$ and $\beta$ did't ocure above then $\Gamma \vDash (\forall \alpha)\phi$
And I don't know how to prove it. 
 A: Hint
Ref to :


*

*Benson Mates, Elementary Logic (2nd edition - 1972).


Assume not, i.e. $\Gamma \nvDash (\forall \alpha)\phi$.
This means that there is an interpretation $\mathfrak I$ satisfying all sentences of $\Gamma$ which does not satisfy $(\forall \alpha)\phi$.
We need the details of the "semantical specifications" used in the textbook : see page 60. 
Saying that $\mathfrak I \nvDash (\forall \alpha)\phi$ means that for some constant $\beta$ :

$\mathfrak I' \nvDash \phi(\alpha / \beta)$

for some $\beta$-variant of $\mathfrak I$.
This contradicts the fact that $\mathfrak I \vDash \phi(\alpha/ \beta)$.
Why so ? Here is the "tricky point" of the proof that needs the proviso : $\beta$ does not occur in any sentences in $\Gamma$.
We have to reacall that a $\beta$-variant of $\mathfrak I$ differ from $\mathfrak I$ only on the value assigned to $\beta$; but $\beta$ does not occur in $\Gamma$.
Thus, if $\mathfrak I$ satisfy $\Gamma$, also any $\beta$-variant of $\mathfrak I$ must satisfy $\Gamma$.
