prove or disprove logical inference I need to prove or disprove the logical inference of the formula b from formula a:
$$ a = \exists y (\forall x P(x) \implies Q(y)) $$
$$ b = \forall x (P(x) \implies \exists y Q(y)) $$
$$ P(x) \space and \space Q(x) \space are \space single-place \space  ratios  $$
Meaning , to show either that every model that satisfy a also satisfy b or to show that there is a model that satisfy a but not satisfy b.
By intuition , I think that I need to prove this , meaning every model that works for a will word for b but I don't know how to approach this.
 A: Let the domain be $\mathbb{N}$, let $P(x)$ be the statement that $x$ is prime, and let $Q(y)$ be the statement that $y \notin \mathbb{N}$.
Then the statement $a$ is clearly true, since the statement $\forall x P(x)$ is false.
Consider that $$\neg b =\neg \forall x [P(x) \Longrightarrow \exists y Q(y)]$$ is equivalent to $$\exists x P(x) \ \& \ \forall y (\neg Q(y)),$$ which is true.
This model provides a counterexample to the statement that $$a \Longrightarrow b$$ is a tautology.
A: $a$ can be rewritten using prenex normal form (nonempty universe assumption), then unprenexing it:
$$\begin{array} {rl}
%
a   &= \exists y ~(\forall x~ Px \implies Qy) 
%
\\  &= \exists y ~\exists x~(Px \implies Qy)
%
\\  &= \exists x ~\exists y~(Px \implies Qy)
%
\\  &= \exists x ~(Px \implies \exists y ~ Qy)
%
\\  &= \forall x ~Px \implies \exists y ~ Qy
%
\end{array}$$
$b$ can be treated similarly:
$$\begin{array} {rl}
%
b   &= \forall x ~(Px \implies \exists y ~ Qy) 
%
\\  &= \exists x ~ Px \implies \exists y ~ Qy
%
\end{array}$$
So the question becomes determining whether:
$$\forall x ~Px \implies \exists y ~ Qy \vDash \exists x ~ Px \implies \exists y ~ Qy$$
So the counter examples are exactly those when $\exists y ~Qy$ is false, and $\exists x ~ Px$ is true, and $\forall x ~ Px$ is false.
