$\exists x\ (P(x) \rightarrow \forall x \ \ P(x))$
For table proof (or tableau proof) could someone let me know what the first step is, I'm confused where to go from here.
$ (1) \ \neg \ \exists x\ (P(x) \rightarrow \forall x \ \ P(x)) \\ (2) \ \neg \ (P(a) \rightarrow \forall x \ \ P(x)) \\ (3) \ P(a) \\ (4) \ \neg \ \forall x \ P(x) \\ (5) \ \neg \ P(b) $
is (2) even correct? Thanks in advanced!
Here's an outline of the proof, as I'm lazy and I'll leave it to you to do the laborious work of actually translating it into proper notation. Assuming we're working in classical logic, we can use the law of excluded middle on ∀ (). The goal then is to use disjunction elimination to get the result. We consider both disjuncts:
If ∀ (), then we get the desired conclusion vacuously (making sure to be careful with introducing the quantifiers) since an implication with a true consequent is always true.
If ¬∀ (), then we use the fact that ¬∀ () is equivalent to ∃¬P(x). Using the counterexample constant for which P does not hold, we obtain the desired result since implication with a false antecedent is always true (again being careful to handle the quantifiers properly).
As an aside, I think the conclusion gives a fun insight. This "worst constant" (which makes P(x) true for all x) does not necessarily exist as an identifiable constant. My professor explained it to me with the statement "There exists a student in this class, who if they get an A, then everyone in the class gets an A." It's true, and you might think, "okay, time to find this student and really force them to study." But there is no such identifiable student! The statement is more about how the nature of implication symbol in classical logic, and how it allows for this sort of non-constructive proof.
