$A \implies B$ in this implication we can say this is false, if we can show a situation where A is true but at the same time B is false.
In the 4th case : $(\exists (x)) (P(x) \vee Q(x)) \implies \sim (\forall (x)) P(x) \vee (\exists (x)) Q(x)$
To check validity we do the following :
- Assume LHS as true or (1) based on some condition.
- Then try to make RHS false (0) on the same set of conditions as above.
- Assume RHS as false (0) and try to make LHS as true (1) on some common conditions.
if we can show anyone one of the above situation exists. Then implication 4 is false.
Using second approach:
in RHS assume all $Q(x)$ are false (0) and all $P(x)$ are true (1) such that RHS becomes false (0).
Based on these above condition if we look on the left hand side, we find that LHS side is true (1) since all $P(x)$ are true.