Proving that ∼(∃x) P (x), is false is equivalent to proving that (∀x )∼P (x) is true.

Question

I found this phrase in the page 60 of the book "A Transition to Advanced Mathematics, 8th Edition, written by Smith/Eggen/St. Andre."

"Proving that ∼(∃x) P (x), is false is equivalent to proving that (∀x )∼P (x) is true."

It seems incorrect to me, but as I am a student, I guess that I am wrong.

My reasoning is that if we work with the equation of the right side:

(∀x)∼P(x) is true, we can see it as

("(∀x)∼P (x) = true") and this would be equal to

("∼ ((∀x)∼P (x)) = ∼(true)")=

(∃x)P(x) = false.

then (∃x)P(x) is false.

That is the opposite of our goal.

Am I wrong?

Thanks!

Answer 1

See page 56 of seventh edition :

There are times when we will want to prove a quantified statement is false. We know that $(\forall x)P(x)$ is false precisely when $\lnot (\forall x)P(x)$ is true and $\lnot (\forall x)P(x)$ is equivalent to $(\exists x) \lnot P(x)$. Therefore, one way to prove $(\forall x)P(x)$ is false is to prove $(\exists x) \lnot P(x)$ is true.

Thus, if $\lnot (\exists x) P(x)$ is false, its negation : $(\exists x) P(x)$ must be true; but $(\exists x) P(x)$ is equivalent to $\lnot (\forall x) \lnot P(x)$.

In conclusion, you are right; alternatively, we can say that :

Proving that $\lnot (∃x)P(x)$, is false is equivalent to proving that $\lnot (∀x) \lnot P(x)$ is true.

As per comments above, the statement in the 2014 edition must be a typo.