Help to prove this formula (Sequent calculus for predicate logic)

Question

I have a formula:

$$¬(\exists x)ϕ(x) ⇒ (∀x)¬ϕ(x) $$

to prove.

If the "$(∃x)ϕ(x)$" was in brackets like this $¬((∃x)ϕ(x))$, I could easily prove this formula, but without it I'm stucked. Can you help me, how can I prove this formul?

Answer 1

Your formula is $\neg((\exists x)\phi(x))$. Negation in front of the quantifier means negation of the quantified statement. The first way of writing it just omits the (redundant) brackets after the negation: $\neg (\phi)$ can be written as $\neg \phi$, so $\neg((\exists x) \phi(x))$ can be written as $\neg (\exists x) \phi(x)$, which was done here. The negation has scope over the entire formula $(\exists x) \phi(x)$, and since the existential quantifier is the outmost operator in the formula, semantically it is indeed the existence is negated.

So the answer is: You'd prove $\neg(\exists x) \phi(x) \to (\forall x) \neg \phi(x)$ in just the same way you'd prove $\neg((\exists x) \phi(x)) \to (\forall x) \neg \phi(x)$ - because they are the same formula.

I would recommend you to have another close look at the definition of the syntax of formulas of predicate logic in the textbook you're using if bracketing is not quite clear yet.