Why does $\exists \ y \ \forall\ x\ Q(x, y) \implies \forall\ x\ \exists\ y\ Q(x, y)$?

Question

Why does $\exists \ y \ \forall\ x\ Q(x, y) \implies \forall\ x\ \exists\ y\ Q(x, y)$? Is it possible to prove this?

Answer 1

1) $\exists y\forall xQ(x,y)$.

2) $\forall xQ(x,b)$.

3) $Q(x,b)$.

4) $\exists y Q(x,y)$.

5) $\forall x\exists yQ(x,y)$.

Answer 2

Imagine an infinitely sized chessboard.

$\exists y \forall x ~Qxy$ says "at least one row of the board is completely inhabited".

$\forall x \exists y ~Qxy$ says "in each column, there is at least one inhabited spot".

The first implies the second because if there is a completely inhabited row, then each column must have at least 1 inhabited spot, that point where the column crosses the row.

Answer 3

Why not proving by contradiction?

Assume this implication doesn't hold, hence $\exists y \forall x Q(x,y)$ and $\lnot \forall x \exists y Q(x,y)$.

Then note that $\lnot\forall x \exists y Q(x,y) \equiv \exists x \forall y \lnot Q(x,y)$. Define $x=:a$ for some element. Then we have $\forall y \lnot Q(a,y)$, in particular $\lnot Q(a,b)$. On the other hand we have $\exists y \forall xQ(x,y)$, we call this $y=:b$. We get $\forall x Q(x,b)$, in particular $Q(a,b)$.

So we have $Q(a,b)$ and $\lnot Q(a,b)$ at the same time, hence our assumption must be wrong.