Predicate Logic: $ ∀x (P(x) \lor Q(x)) ⊢ ∀xP(x) \lor ∀xQ(x) $ The exercise ask me to prove it using natural deduction or show a example that false the sequent. How do i prove it using natural deduction ?
$ \forall x (P(x) \lor Q(x)) ⊢ ∀xP(x) \lor \forall xQ(x) $
I tried:
1.$ ∀x (P(x) \lor Q(x)) $


*$ x_0 $

*$ P(x_0) V Q(x_0) $ 


Then I tried to use the or-exclusion rule but i can not figure out how to get $P(x_0)$ from $Q(x_0)$
 A: This isn't true. 
Let $P(x)$ be "x is a banana" and $Q(x)$ be "x isn't a banana" and $x$ be from the space of all fruit.
Then the predicate is true because every fruit either is or is not a banana.
It doesn't follow that either all fruits are bananas or all fruits are not bananas.
So for a counter example simply choose $P(x) = \lnot Q(x)$. And a space for which some elements $P(x)$ is true and other elements $P(x)$ is false.
A: What you'll find is that you cannot prove the statement.  It is false.  Whenever you're hitting dead ends in a proof, try considering a counterexample. 
Let's look at the outcomes from flipping a coin: $P(x)$: "x is heads," $\;\;Q(x)$: "x is tails", with x belonging to the domain of all coins.
Then we can say truthfully that $$\forall x (P(x)\lor Q(x)).$$
But there's a problem with the RHS: It's claiming that $$\forall x (P(x)) \lor \forall x (Q(x))$$
That is, the right hand side claims that every coin tossed turns of heads, or else, every coin tossed turns of tails.
