Argthtjtr

I know that the rules for removing quantifiers from the antecedent of a conditional are as follows:
(∀x)φ(x) → ψ ⇔ (∃x)(φ(x) → ψ) and (∃x)φ(x) → ψ ⇔ (∀x)(φ(x) → ψ), provided that x is not free in ψ.
I also know that the rules for removing quantifiers from the consequent are as follows:
φ → (∀x)ψ(x) ⇔ (∀x)(φ → ψ(x)) and φ → (∃x)ψ(x) ⇔ (∃x)(φ → ψ(x)), provided that x is not free in φ.

Now, I understand how to derive these equivalences syntactically. However, I can't seem to grasp how the first two can be semantically equivalent.

For example, how can two sentences such as "If all x are wet, then it's raining" and "There exists an x such that if x is wet, then it's raining" can be equivalent?

Moreover, provided that x is not free in ψ, isn't (∀x)φ(x) → ψ equivalent to (∀x)(φ(x) → ψ)? Which would result in the absurd equivalence: (∀x)(φ(x) → ψ) ⇔ (∃x)(φ(x) → ψ).

I think that my difficulty here arises from a misunderstanding of the significance of the parentheses in the aforementioned equivalences. Please provide examples in your answer if you can.

how can two sentences such as "(If all x are wet, then it's raining)" and "There exists an x such that (if x is wet, then it's raining)" can be equivalent?

A simple approach to this equivalence, that holds only in classical logic, is to exploit the so-called Material implication rule :

(P → Q) ≡ (¬ P ∨ Q).

Thus, we have that : "(If all x are wet, then it's raining)" is equivalent to : "Either (not all x are wet) or it's raining", i.e "Either (there exists an x which is not wet) or it's raining".

Now we have to use the fact that the existential quantifier "distribute" over or and the fact that "it's is raining" is equivalent to "there is an x such that it's raining", due to the key-fact that x is not free in "it's raining".

Thus, the last step is to move the existential quantifier in front of the formula to get :

"There exists an x such that (either x is not wet or it's raining)."

We can show the semantical equivalence of :

(∀x βx → α) ↔ ∃x(βx → α)

this way.

Let M a structure whatever, with domain D.

Two cases :

(i) ∀x βx is False. In this case (∀x βx → α) is True.

This means that there is an a in D such that βa is False.

Thus, (βa → α) is True, which means that (βx → α) is True for some x.

And thus, also ∃x(βx → α) is True.

(ii) ∀x βx is True.

Two subcases :

(iia) α is True. In this case (∀x βx → α) is True.

But if α is True, also (βx → α), irrespective of the value of x, and thus also ∃x(βx → α) is True.

(iib) α is False. In this case (∀x βx → α) is False.

The fact that ∀x βx is True imples that βa is True, for every a in D.

Thus (βa → α) is False, for every a in D, which means that ∃x(βx → α) is False.