Proof about distribution of $\forall$ universal quantifier over implication

Question

I am struggling with this proof:

$\forall x(A \to B) \vDash A \to \forall x B$, if $x \not \in {free}(A)$

Below is my thoughts:

Proof: Let $\mathfrak{I} =(D,\varphi)$ be an interpretation, and let $\sigma$ be a variable assignment over $\mathfrak{I}$.

Assume $\mathfrak{I}, \sigma \vDash \forall x(A \to B), $ so for all d $\in$ D, $\mathfrak{I} , \sigma^{\frac{d}{x}} \vDash A \to B $,

so either $\sigma \nvDash A $ or $\sigma \vDash B$,

if $\sigma \nvDash A $, then $\sigma \vDash A \to \forall xB$, then I have no idea how to do...

I want to know how to prove this claim. Would anyone help me out please?

Answer 1

By contradiction: assume that the premise holds and the conclusion does not, i.e. that:

$I, \sigma \vDash \forall x (A \to B)$ and $I, \sigma \nvDash A \to \forall xB$.

This means that $I, \sigma \vDash A$ and $I,\sigma \nvDash \forall x B$.

This implies that, for some $d \in D$, we have that $I, \sigma (d|x) \nvDash B$.

But due to the fact that $x \notin \text {FV}(A)$, we have that form $I, \sigma \vDash A$, also: $I,\sigma(d|x) \vDash A$.

In fact, if $x$ does not occur free in $A$, and $\sigma$ satisfies $A$, also the new sequence $\sigma(d|x)$ that assigns $d$ to $x$ will satisfy it.

And thus:

$I, \sigma (d|x) \nvDash (A \to B)$,

contradicting the assumption above.