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.

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

And vice-versa.


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