# Understanding the proof of $\sim\!\!(\forall x)A(x)\!\iff\!(\exists x)\sim \!\! A(x)$

I need to understand the proof of $$\sim\!\!(\forall x)A(x)\!\iff\!(\exists x)\sim \!\! A(x)$$ in "A Transition to Advanced Mathematics: Edition 8" pg. 23 Theorem 1.3.1 a)

It states:

1. Let $$U$$ be any universe (universe of discourse, the set of values being considered for $$x$$)
2. The sentence $$\sim\!\!(\forall x)A(x)$$ is true in U
• iff the truth set of $$A(x)$$ is not the universe
• iff the truth set of $$\sim A(x)$$ is nonempty
• iff $$(\exists x)\sim \!\! A(x)$$ is true in U

I couldn't rationalize how "iff the truth set of $$\sim A(x)$$ is nonempty" led to " iff $$(\exists x)\sim \!\! A(x)$$ is true in U". Couldn't we also say "iff $$(\forall x)\sim \!\! A(x)$$ is true in U"?

Is there a better proof I can understand? So far I learned about negation, conjuctions, disjunctions, conditionals and biconditionals.

• not all mammals are dogs is equivalent to there are mammals that are not dogs, not to all mammals are not dogs – J. W. Tanner Jan 20 at 16:29
• btw, you can use \lnot for $\lnot$. – Dave Jan 20 at 16:43

$$\lnot(∀x)A(x)$$ means that in the "universe" $$\text U$$ not every object is an $$A$$.
In the "universe" $$\mathbb N$$ of natural numbers, not every number is even.
Thus, there is some object that is a not-$$A$$, i.e. $$(∃x)\lnot A(x)$$.
In $$\mathbb N$$ there are numbers that are not-even, i.e. odd.