Validity of ∃x(Qx→∀x Qx ).

Question

I was reading through Enderton's "Mathematical Introduction to Logic" and came across 2 examples, one a formula, one a relationship:

1. ∃x(Qx→∀x Qx)

2. ∀y∃x Pxy $\nvDash$ ∃x∀yPxy

Enderton claims the first one is a validity. I fail to see how this is the case. How would I show this?

For the second one, I explicitly defined a predicate P: a 2 place predicate that says ($y*x = 1)$ so the statement would read as following: For all y, there exists such that ($y*x = 1)$ but the right side would read: there exists x such that for all y, ($y*x = 1)$. Now if you do it the converse way, ie ∃x∀yPxy $\vDash$ ∀y∃x Pxy then this relationship is actually true. My question is, is the example I provided enough to show the relationship holds for (2)?

edit: Not an exact duplicate, since there is a second part of the question that is not answered in the linked thread.

Answer 1

One way to think of (1) is that we can let $y$ be the "least $Q$ish" of all elements. If $y$ is actually $Q$ish, then all elements must be $Q$ish!

For instance if the least tall person, $y$, is actually tall, then all persons $x$ are tall. So $$Qy\to \forall xQ(x)$$ and hence $$\exists y(Qy\to \forall xQ(x)).$$

Answer 2

The second part is trivial.

In order to show that:

$∀y∃x Pxy \nvDash ∃x∀yPxy$,

consider the domain $\mathbb N$ and interpret $P$ with $\ge$ (i.e. $P^{\mathbb N} = \ge$).

Clearly: $∀m∃n \ (n \ge m)$ holds but $∃n∀m (n \ge m)$ does not (i.e. $\mathbb N \vDash ∀m∃n \ (n \ge m)$ and $\mathbb N \nvDash ∃n∀m (n \ge m)$).