Discrete Mathematics: Predicate Logic

Question

Is the following implication valid?

$(∃(x))(P(x)∨Q(x))⟹\lnot (∀(x))P(x)∨(∃(x))Q(x)$

For proving this I used to follow a method to by making L.H.S as 1 and try to make R.H.S as 0.

My work:-

I am having a small trouble while reading the R.H.S of the implication.The scope of the negation is bounded to only Universal quantifier or to the entire statement?

If the R.H.S is :- $\lnot ∀(x)P(x)∨(∃(x))Q(x)$,then it means the negation of the entire statement.But as there are braces, I am getting a bit confused.As I have read that Quantifiers has higher precedence, does this mean it is the negation of only the quantifier?

How should I read it:- $(∃(x))P(x)∨(∃(x))Q(x)$ or

$(∃(x))\lnot P(x)∨(∃(x))Q(x)$

Answer 1

The left hand side reads: 'There is something that is either a $P$ or a $Q$'. The right hand side reads: 'Either not everything is a $P$, or there is something that is a $Q$.

There is a simple counterexample to this implication: consider a domain with just one object, that has property $P$, but not $Q$. Then there is something that is either a $P$ or a $Q$ (since it is a $P$), so the left hand side is True. But it is not true that not everything is a $P$ (since everything is a $P$), or that there is smething that is a $Q$, and hence the right hand side is false. So, the implication does not hold.

Answer 2

(∃(x))(P(x)∨Q(x))⟹∼(∀(x))P(x)∨(∃(x))Q(x)

Removing uneccessary brackets, adding spacing to improve readability, and highlighting the scope of each quantifier, this is:

$$\exists x~\bbox[1pt,lemonchifon,border:solid pink 0.1pt]{\big(P(x)\vee Q(x)\big)}~\implies~\neg \forall x~\bbox[1pt,lemonchifon,border:solid pink 0.1pt]{P(x)}~\vee~\exists x~\bbox[1pt,lemonchifon,border:solid pink 0.1pt]{Q(x)}$$

Thus reading: "There is something that is P or Q, implies that not everything is P, or something is Q."

Now, consider the universe of one element, $\{a\}$, where $P(a)$ but not $Q(a)$.