Predicate Logic Equivalence Intuition

Question

Feel free to redirect me if this was asked before. I have troubles understanding the following identity of predicate logic (it holds only if the domain of discourse is non empty): $$ \forall x \phi(x) \Rightarrow C \equiv \exists x(\phi(x) \Rightarrow C) $$ In words: If all $x$ satisfy $\phi$ then C is true $\equiv$ there exists at least one $x$ so that if $x$ satisfies $\phi$ then C is true.

For example let

• $x$: question in exam
• $\phi(x)$: $x$ was solved
• $C$: i passed the exam

Then the left side of the equivalency reads "If i solved every question in the exam then i passed the exam". The right side reads "There exists an exercise that, if solved correctly, makes me pass the exam. That does not seem right. If i have 3 questions with 5 points each and i need 8 points to pass, no question on its own directly implies passing if solved correctly. Do you see where i made a mistake? And could you formulate the right version of my example? Thank you very much in advance.

Answer 1

$$(\forall x \phi(x) \Rightarrow C) \equiv \exists x(\phi(x) \Rightarrow C)\tag1$$

• $x$: question in exam
• $\phi(x)$: $x$ was solved
• $C$: i passed the exam

The left side of the equivalency reads "If I solved every question in the exam then I passed the exam".

The right side reads "There exists an exercise that, if solved correctly, makes me pass the exam.

Your left- and right-side translations aren't consistent with each other: the right side should instead be "for some question, if I solved it, then I passed the exam". On both sides, we're describing a past event, perhaps due to incomplete knowledge rather than the marking rubric's specifications.

That does not seem right. If i have 3 questions with 5 points each and I need 8 points to pass, no question on its own directly implies passing if solved correctly.

By correctly noting that $$(\forall x \phi(x) \Rightarrow C) \kern.6em\not\kern-.6em\implies (\exists x\phi(x) \Rightarrow C),$$ you are countering the false claim $$(\forall x \phi(x) \Rightarrow C) \equiv (\exists x\phi(x) \Rightarrow C),\tag2$$ which is a stronger claim than statement $(1).$

The forward direction of statement $(1)$ is troubling you, so let's prove it. Suppose that its LHS premise $(\forall x \phi(x) \Rightarrow C)$ is true; then either of these possibilities must be true:

1. passed the test:

   then $(\phi(x) \Rightarrow C)$ is true, so $\exists x(\phi(x) \Rightarrow C)$ is true.

2. failed the test:

   then, by letting $x=p$ be the missed question, $\exists x(\phi(x) \Rightarrow C)$ is vacuously true.

Answer 2

$\exists x(Px\implies C)$

$\iff\exists x(\neg Px \lor C)$

$\iff\exists x(\neg Px) \lor C$

$\iff\neg\forall x(Px) \lor C$

$\iff \forall x(Px)\implies C$

The problem with the second predicate in your question is that, an implication inside of an existential quantifier simply does not have the same "intuitive" meaning that it does inside of a universal one. In these cases, just treat it as a disjunction (which it is by definition).