How to solve this logics problem?

Question

question is that

\begin{align} 1.∀ x (P(x)⟹Q(x)∨R(x))\\ 2.∀ x (P(x)∧ㄱR(x))\end{align} hence $∃xQ(x)$

i know that $$∀x(p(x)∨q(x))≢∀x(p(x))∨∀x(q(x))$$

then $$∀xP(x)≢∀xQ(x)∨ ∀xR(x)$$ right?

Hmm I don't understand how to solve this problem.

Answer 1

There exists such $x$, provided that some $x$ exists... That is, if the space under consideration is nonempty.

Otherwise both given conditions are vacuously true, anyway still no $x$ exists.

As for the proof:

by 1. we have for all $x$ the implication holds:

$$\forall x \Big( P(x)\implies Q(x)\lor R(x)\Big)$$

Now we can simplify the whole expression a bit. Let's start from the implication equivalence first: $$(A\implies B)\equiv (\lnot A \lor B)$$

Then (assuming we have '(Q(x) ∨ R(x))' in the implication in 1.) it follows that $$\forall x \Big( \lnot P(x)\lor (Q(x)\lor R(x)) \Big)$$ $$\forall x \Big( Q(x)\lor \color{green}{(\lnot P(x)\lor R(x))} \Big)$$ $$\forall x \Big( Q(x)\lor \lnot \color{green}{(P(x)\land \lnot R(x))} \Big)$$

BUT...

by 2. the internal, green subexpression is satisfied ($TRUE$) for all $x$, so:

$$\forall x \Big( Q(x)\lor \lnot \color{green}{TRUE} \Big)$$ $$\forall x \Big( Q(x)\lor \color{green}{FALSE} \Big) $$ $$\forall x \ Q(x)$$

Hence, if there exists any $x$, then there exists $x$ which satisfies $Q(x)$.