Can universal generalization be used on logical equivalence?

Question

I'm a high schooler studying logic from a few discrete math books and I wish to show that $\forall x, (P(x) \rightarrow Q(x))$ is logically equivalent to $\forall x, (\neg Q(x) \rightarrow \neg P(x)),$ using universal generalization to show this logical equivalence.

I'm aware that to prove logical equivalence for quantified statements, I have to show that the biconditional between the two quantified statements is a tautology. So I'll have to first show that the conditional statement $$(\forall x, (P(x) \rightarrow Q(x))) \rightarrow (\forall x, (\neg Q(x) \rightarrow \neg P(x)))$$ is a tautology, then show that the converse of it is also a tautology.

But by using universal generalization, I feel like I could skip this and just prove that they're logically equivalent without needing to go through conditional statements but I'm not sure if this is the correct way to do it. Here is how the proof goes:

Suppose that $\forall x, (P(x) \rightarrow Q(x))$ is true, then whenever $P(x)$ is true, $Q(x)$ will also be true. Let $a$ be any arbitrary element from the domain, then $P(a) \rightarrow Q(a)$ is a true conditional statement. Since a conditional statement is logically equivalent to its contrapositive (proved by using truth table for statements), we have $$P(a) \rightarrow Q(a) \equiv \neg Q(a) \rightarrow \neg P(a)$$ And since $a$ was arbitrary, we can use universal generalization to generalize it to $$\forall x, (P(x) \rightarrow Q(x)) \equiv \forall x, (\neg Q(x) \rightarrow \neg P(x))$$

As you can see, I didn't show if the conditional statements were tautologies. Am I using universal generalization correctly here? Does this make my proof correct? If not, how will I need to correct my proof? I'm not so sure if I can use universal generalization to go from $$P(a) \rightarrow Q(a) \equiv \neg Q(a) \rightarrow \neg P(a)$$ to $$\forall x, (P(x) \rightarrow Q(x)) \equiv \forall x, (\neg Q(x) \rightarrow \neg P(x)).$$

I'm using the symbol ≡ in the metalanguage sense instead of as the material biconditional. I use the symbol ↔ to mean the material biconditional.

Answer 1

Suppose that $\forall x\, (P(x) \rightarrow Q(x))$ is true

Let $a$ be any arbitrary element from the domain; then $P(a) \rightarrow Q(a)$

Since a conditional statement is logically equivalent to its contrapositive (proved by using truth table for statements), we have $$P(a) \rightarrow Q(a) \equiv \neg Q(a) \rightarrow \neg P(a)$$

This last line in symbols—which is merely a rephrasing of your explanation preceding it— can be used to derive the tautological consequence $$\neg Q(a) \rightarrow \neg P(a);$$ we then apply Universal Generalisation to derive the required RHS, giving $$\forall x\, (P(x) \rightarrow Q(x)) \vdash \forall x\, (\lnot Q(x) \rightarrow \lnot P(x)).$$

And since $a$ was arbitrary, we can use universal generalization to generalize it to $$\forall x\, (P(x) \rightarrow Q(x)) \equiv \forall x\, (\neg Q(x) \rightarrow \neg P(x))$$

Now, writing neither the metalogical assertion $$P(a) \rightarrow Q(a) \equiv\neg Q(a) \rightarrow \neg P(a)$$ nor the tautological consequence $$(P(a) \rightarrow Q(a)) \leftrightarrow (\neg Q(a) \rightarrow \neg P(a))$$ amounts to asserting that either of the two conditionals is true (i.e., actually deriving either of the two conditionals). Applying Universal Generalisation to the latter gives $$\forall x\, (P(x) \rightarrow Q(x)) \vdash \forall x\,\big( (P(x) \rightarrow Q(x)) \leftrightarrow (\neg Q(x) \rightarrow \neg P(x))\big),$$ which isn't what we want.

I'm not so sure if I can use universal generalization to go from $$P(a) \rightarrow Q(a) \equiv \neg Q(a) \rightarrow \neg P(a)$$ to $$\forall x (P(x) \rightarrow Q(x)) \equiv \forall x (\neg Q(x) \rightarrow \neg P(x)).$$ I'm using the symbol ≡ in the metalanguage sense instead of as the material biconditional. I use the symbol ↔ to mean the material biconditional.

It doesn't make sense during a sequence of syntactical derivations to suddenly teleport to the metalanguage to apply Universal Generalisation—a syntactical rule that transforms sentences in the object language—there.

Nor does it make sense to claim that Universal Generalisation transforms $$M(a)\leftrightarrow N(a)\tag1$$ to $$\forall x\;M(x)\leftrightarrow \forall x\;N(x),$$ since its rule explicitly says to append the quantifier around the entirety of formula $(1).$

Furthermore (rephrasing my parenthetical comment above), remember, formula $(1)$ does not mean $$M(a)\land N(a);$$ and Universal Generalisation per se does not transform even this formula to $$\forall x\;M(x)\land \forall x\;N(x).$$ P.S. For reference: $$\forall x\;M(x)\land \forall x\;N(x)\quad\equiv\quad \forall x\;\big(M(x)\land N(x)\big)\\ \forall x\;M(x)\leftrightarrow\forall x\;N(x)\quad\not\equiv\quad \forall x\;\big(M(x)\leftrightarrow N(x)\big).$$