show that ∀ x P ( x ) ∨ ∀ x Q ( x ) is logically equivalent to ∀ x ∀ y ( P ( x ) ∨ Q ( y )) . (Domains for x and y are the same). My attempt at a solution:

Proof that $\forall xP(x)\vee\forall xQ(x)\equiv\forall x\forall y(P(x)\vee Q(y))$: 
  
  
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*Suppose $\forall xP(x)\vee\forall x Q(x)$ is true. Then $P(x)$ is true for all $x$ or $Q(x)$ is true for all $x$. Since the domain of $x$ and $y$ is the same, $P(x)$ is true for all $x$ or $Q(y)$ is true for all $y$. So, for all $x$ and $y$, either $P(x)$ is true or $Q(y)$ is true. Then $\forall x\forall y (P(x)\vee Q(y))$ is true.
  
*Suppose $\forall xP(x)\vee\forall xQ(x)$ is false; Then there exists an $x_0$ such that $P(x_0)$ is false and a $y_0$ such that $Q(y_0)$ is false. Then $P(x)$ is false for all $x$ and $Q(y)$ is false for all $y$. Therefore, $\forall x\forall y(P(x)\vee Q(y))$ is false.
  

 A: In general you do a too complicated proof by trying to use contradiction proofs. Assume $a,b$ are any elements in the domain (we need to assume this in order to prove $\forall$ quantifiers).
If $\forall x P(x)\vee \forall x Q(x)$ hold,  then assume $\forall xP(x)$ hold (parallel proof if $\forall x Q(x)$ hold). In this case we know that $P(a)$ hold, as $a$ is an element in the domain, but then also $P(a)\vee P(b)$ hold. As $a$ and $b$ are general elements we may thus conclude that $\forall x\forall y(P(x)\vee P(y))$ hold.
If $\forall x\forall y(P(x)\vee P(y))$ is true then $P(a)\vee P(b)$ hold, as $a$ and $b$ are in the domain. If $P(a)$ hold then, since $a$ is a general element, we may conclude that $\forall x P(x)$ hold, and thus $\forall xP(x) \vee \forall x Q(x)$ hold.
A: This is a special case of $A \lor \forall x ~ B(x) \equiv \forall x ~ A \lor B(x)$.  Split cases:
Case 1: $A$ is true:
$$\text{true} \lor \forall x ~ B(x) \equiv \forall x ~ \text{true} \lor B(x)$$
  $$\text{true} \equiv \forall x ~ \text{true}$$
  $$\text{true}$$
Case 2: $A$ is false:
$$\text{false} \lor \forall x ~ B(x) \equiv \forall x ~ \text{false} \lor B(x)$$
  $$\forall x ~ B(x) \equiv \forall x ~ B(x)$$
  $$\text{true}$$
You have to apply the theorem twice:
$$\forall x~P(x) \lor \forall y ~Q(y)$$
Apply with $A = \forall y ~ Q(y)$ and $B = P$:
$$\forall x ~(P(x) \lor \forall y~Q(y))$$
Apply with $A = P(x)$ and $B = Q$ :
$$\forall x ~ \forall y~(P(x) \lor Q(y))$$
