Why is this expression false? I have thought thoroughly over this expression $(\:∀x∃y\:(Q(x, y) ⇒ P(x))\:) ≡ (\:∀x\:(∃y\:Q(x, y) ⇒ P(x))\:) $ however I do not understand why they are not equivalent? I thought the "there exists" symbol distributes because P(x) does not use it? Similarly, why is $(\:∀x∃y\:(P(x) ⇒ Q(x,  y))\:) ≡ (\:∀x\:(P(x) ⇒ (∃y\:Q(x, y)))\:) $ true? Because the forall distributes across nonusers?

Sorry about the confusion, the expressions have been updated and finalized.
Remember that $Q(x,y)\to P(x)$ is equivalent to $\neg Q(x,y)\vee P(x)$

So when you distribute the existential over the implication into the antecedent, you have to convert to the dual (that is a universal).

$$\begin{align}\forall x\;\exists y\;\big(Q(x,y)\;\to\; P(x)\big) &amp; \equiv \forall x\; \exists y\;\big(\neg Q(x,y)\;\vee\; P(x)\big) \\[1ex] ~ &amp; \equiv \forall x \;\Big(\big(\exists y\;\neg Q(x,y)\big)\;\vee\; P(x)\Big) \\[1ex] ~ &amp; \equiv \forall x\;\Big(\neg\big(\forall y\; Q(x,y)\big) \;\vee\; P(x)\Big) \\[1ex] ~ &amp; \equiv \forall x\;\Big(\big(\forall y\;Q(x,y)\big)\;\to\; P(x)\Big)\end{align}$$
There is a parenthesis after P(x) in RHS. For all distributes over both sides entirely (both over P and Q) for the first expression.