Doubt in predicate calculus equivalence: $\forall x P(x) \rightarrow Q ≡ \exists x (P(x) \rightarrow Q))$

Question

I am learning predicate calculus. I came across this identity. $$(\forall x) P(x) \rightarrow Q ≡ \exists x (P(x) \rightarrow Q)$$ Since then I have been scratching my head. Algebraically, it can be very easily shown that the above holds but I am not able to develop correct intution for it. I took the following example to understand it better. But rather than helping me, I am more confused now.

Example

P(x): x scores more than 90%

Q: School management throws a party

Then, $(∀x)P(x)→Q$ would mean that "if all students score more than 90%, then school management will through a party."

$\exists x (P(x) \rightarrow Q)$ would mean that there exists at least one student such that if he/she scores above 90%, then school management will give party.

Both don't seem to match in their meaning. Can someone please correct me.

Question

I want help in developing intution about this identity. For this, I request you to please correct me in the example that I gave. Probably by the help of it, I will get a better idea of the identity.

Answer 1

As requested in a comment, I'm promoting my previous comment to an answer:

The "one student" in the $\exists$ formulation is the student with the lowest score. If that looks like cheating then you should review the definition of what it means for a sentence to be true in a structure, in particular the $\exists$ clause in that definition. It allows the witness (the $x$ in your example) to change when the structure (which includes the students' scores) changes.

Answer 2

Using your example, the following is an intuitive exlplanation of $\forall x [Px] \to Q \vdash \exists x [Px \to Q]$.

Assume $\forall x [Px] \to Q$. In other words, assume that if every student scores $> 90$%, then school management throws a party. Now, there are two cases:

(i) every student scores $> 90$%

(ii) at least one student scores $\le 90$%

case (i). Suppose every student scores $> 90$%. Then, by Modus Ponens with $\forall x [Px] \to Q$, we may conclude that school management throws a party. Then, for any particular student, say student $a$, it cannot be wrong to say that if student $a$ scores $> 90$%, then school management throws a party - because we know that every student scores $> 90$% and we have already determined that school management throws a party. If every student scores $> 90$%, then so does student $a$ because "every student" includes student $a$. Thus, in the case where every student scores $> 90$%, there exists at least one student such that, if that student scores $> 90$%, then school management throws a party.

case (ii). Suppose at least one student scores $\le 90$%. Let that student be student $b$. Is it not true that if student $b$ scores $> 90$%, then school management throws a party? Since student $b$ does not score $> 90$%, this implication holds true vacuously. Thus, in the case where at least one student scores $\le 90$%, there indeed exists a student (that is, any student who scores $\le 90$%) such that, if that student scores $> 90$%, then school management throws a party.

So, in summary, given $\forall x [Px] \to Q$, we have two cases: either every student scores $> 90$%, or at least one student scores $\le 90$%. In each case, there exists a student such that, if that student scores $> 90$%, then school management throws a party. Hence, we may conclude $\exists x [Px \to Q]$.

An intuition for $\exists x [Px \to Q] \vdash \forall x [Px] \to Q$ is more straightforward.

Assume $\exists x [Px \to Q]$. In other words, assume there exists at least student such that, if that student scores $> 90$%, then school management throws a party. Let that particular student be student $c$. Then, we know if student $c$ scores $> 90$%, then school management throws a party, or in other words, we know $Pc \to Q$. Now, assume every student scores $> 90$%. Well, "every student" includes student $c$, so it must be the case student $c$ scores $> 90$%. In turn, by Modus Ponens with $Pc \to Q$, we may conclude that school management throws a party. Thus, given $\exists x [Px \to Q]$, we know that if every student scores $> 90$%, then school management throws a party. In other words, we know $\forall x [Px] \to Q$.

Given the proofs above, we may conclude the two formulas $\forall x [Px] \to Q$ and $\exists x [Px \to Q]$ are syntactically equivalent. In other words...

$$\forall x [Px] \to Q \dashv \vdash \exists x [Px \to Q]$$

In order to cultivate an understanding of counterintuitive arguments and statements like these, I have found the following approach helpful: first develop a formal proof using the inference rules of FOL. Then, follow the logic of the proof, step by step, from beginning to end. I have not included any formal proofs in this post because you emphasized the desire for intuition.