Prove $(\forall x_i)(\mathscr{B}\rightarrow \mathscr{C})\rightarrow ((\forall x_i)\mathscr{B}⇒(\forall x_i)\mathscr{C})$ is logically valid

Question

I've begun to show the above as being logically valid like so:

Assume $(\forall x_i)(\mathscr{B}\rightarrow \mathscr{C})\rightarrow ((\forall x_i)\mathscr{B}⇒(\forall x_i)\mathscr{C})$ is not logically valid. Then there is an interpretation $M$ for which it is not true. Hence, there is a sequence $s$ in $\Sigma$ such that s satisfies $(\forall x_i)(\mathscr{B}\rightarrow \mathscr{C})$ and $s$ does not satisfy $((\forall x_i)\mathscr{B}⇒(\forall x_i)\mathscr{C})$. Since $s$ satisfies $(\forall x_i)(\mathscr{B}\rightarrow \mathscr{C})$, s satisfies $(\mathscr{B}\rightarrow \mathscr{C})$.

I get this far but then start to wonder if a reductio is the right action to take. Since, if I were to go on to say that $s$ then satisfies $\lnot((\forall x_i)\mathscr{B}⇒(\forall x_i)\mathscr{C})$, therefore it satisfies $\lnot(\mathscr{B}\rightarrow\mathscr{C})$ I believe I would be making a mistake - or at least missing out some steps.

Can $s$ at this point satisfy either of the quantifiers in the consequent?

Answer 1

See pag.57:

$s$ satisfies $(\forall x_i)(\mathscr{B} \to \mathscr{C})$ if and only if every sequence that differs from $s$ in at most the $i$th component satisfies $(\mathscr{B}\rightarrow \mathscr{C})$.

But $s$ does not satisfy $(∀x_i)\mathscr{B} \to (∀x_i)\mathscr{C}$, i.e. $s$ satisfies $(∀x_i)\mathscr{B}$ and does not satisfy $(∀x_i)\mathscr{C}$.

This means that every $x_i$-variant of $s$ satisfies $\mathscr{B}$ but there is an $x_i$-variant of $s$, call it $s'$, that does not satisy $\mathscr{C}$.

This means that $s'$ satisfies $\mathscr{B}$ but not $\mathscr{C}$, and thus it does not satisfy $\mathscr{B} \to \mathscr{C}$.

Thus, we have found a sequnce $s'$ that differs from $s$ in at most the $i$th component that does not satisfy $\mathscr{B} \to \mathscr{C}$, contradicting the assumption above that every sequence...

Answer 2

There is a syntactic proof of $(\forall x_i)(\mathscr{B}\Rightarrow \mathscr{C})\Rightarrow ((\forall x_i)\mathscr{B} \Rightarrow (\forall x_i)\mathscr{C})$ as follows.

1. $(\forall x_i)(\mathscr{B}\Rightarrow \mathscr{C})$, Hyp
2. $(\forall x_i)\mathscr{B}$, Hyp
3. $\mathscr{B}\Rightarrow \mathscr{C}$, by 1
4. $\mathscr{B}$, by 2
5. $\mathscr{C}$, by 3, 4 and MP rule
6. $(\forall x_i)\mathscr{C}$, by 5 and Gen rule
7. $(\forall x_i)(\mathscr{B}\Rightarrow \mathscr{C}), (\forall x_i)\mathscr{B} \vdash (\forall x_i)\mathscr{C}$, by 1-6
8. $\vdash (\forall x_i)(\mathscr{B}\Rightarrow \mathscr{C})\Rightarrow ((\forall x_i)\mathscr{B} \Rightarrow (\forall x_i)\mathscr{C})$, by applying Deduction Theorem twice since $x_i$ is not free in the two hypotheses.

The above syntactic proof only uses logical axioms, i.e. it is a predicate calculus. In a predicate calculus, any theorem is logically valid because all logical axioms are logically valid and the inference rules preserve logical validity. Therefore, $(\forall x_i)(\mathscr{B}\Rightarrow \mathscr{C})\Rightarrow ((\forall x_i)\mathscr{B} \Rightarrow (\forall x_i)\mathscr{C})$ is logically valid.