Universal Quantification over Disjunction

Question

I know that $\forall x [R(x) \rightarrow S(x)]$ is NOT equivalent to $\forall x R(x) \rightarrow \forall x S(x)$, because universal quantification does not distribute over disjunction. However, it seems to me that $$(*) \forall x [R(x) \rightarrow S(x)] \rightarrow [\forall x R(x) \rightarrow \forall x S(x)]$$ just intuitively. (For context, I am studying downward-entailments in linguistic semantics.).

How can I prove (*)?

Answer 1

Hint: One way in which you can proceed is by using a semantic argument (i.e., by using models) and showing that it is valid in any model that you take. Then, apply the Gödel's completeness theorem.

However, if you want a syntactic proof, you must give us your axioms first, because $(*)$ is given like an axiom in many books.

Answer 2

If you are using natural deduction axioms/rules (e.g. as per Van Dalen), I think the folowing is a proof: (A stands for assumption, Best viewed on a PC; is there a good way of presenting tables in Latex?).

$Level | Line | Formula \quad\quad\quad\quad\quad \quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad |Rule \quad |Depends on\quad | Cancelled by $
$2 \quad\quad 1\quad\quad \quad\quad \forall x (R(x) \rightarrow S(x)) \quad\quad\quad \quad \quad\quad\quad\quad\quad\quad\quad\quad A \quad \quad \quad \quad \quad \quad \quad\quad\quad8$
$2 \quad\quad 2 \quad\quad\quad \quad(R(x) \rightarrow S(x))\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad \forall x E \quad \quad 1$
$3 \quad\quad 3 \quad\quad\quad \quad \quad \quad\forall x R(x)\quad \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad A \quad\quad\quad\quad\quad\quad\quad\quad\quad 7 $
$3 \quad\quad 4 \quad\quad\quad \quad\quad \quad R(x) \quad \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \forall x E \quad \quad 3$
$3 \quad\quad 5 \quad\quad\quad\quad\quad \quad S(x)\quad \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \rightarrow I \quad \quad 2$
$3 \quad\quad 6\quad \quad\quad\quad\quad\quad \forall x S(x)\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \forall x I \quad \quad 5 $
$2 \quad\quad 7\quad\quad\quad \quad (\forall x R(x)\ \rightarrow \forall x S(x))\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \rightarrow E \quad \quad 3,6$
$2 \quad\quad 8\quad \quad (\forall x (R(x) \rightarrow S(x)) \rightarrow (\forall x R(x)\ \rightarrow \forall x S(x)))\quad\quad \rightarrow E \quad\quad 1,7 $

That is, suppose $\forall x (R(x) \rightarrow S(x)) $. Then $R(x) \rightarrow S(x)$ for an arbitrary $x$. Suppose $\forall x R(x)$. Then $R(x)$ and so $S(x)$. As so as $x$ was arbitrary $\forall x S(x)$. Hence $(\forall x R(x)\ \rightarrow \forall x S(x))$.

Alternately, suppose $\forall x (R(x) \rightarrow S(x)) $. Suppose $\forall x R(x)$. Suppose $\forall x S(x)$ is false. Then there is an $x$ such that $S(x)$ is false. And for this $x$, $R(x)$ is true. Hence $(R(x) \rightarrow S(x))$ is false. And so $\forall x (R(x) \rightarrow S(x))$ is false, a contradiction to our initial assumption. Hence $\forall x S(x)$ is true and $(\forall x R(x)\ \rightarrow \forall x S(x))$