Your argument after step 3 is not correct.
The conditional : $(∃z(x=y*z)⇒(y=x)∨(y=1))$ is universally quantified by $∀y$. This means that : for any $y$ we have to "test" if the antecedent $∃z(x=y*z)$ holds.
Thus, for the sake of argument, assume $x = 3$; we have that $x \ne 1$ and thus the first conjunct is true.
For the second conjunct, we have to check, for any $y$, if exists $z$ such that : $3=y*z$.
For $y = 1$ we have $z = 3$; for $y = 2$ there is no $z$; for $y = 3$ we have $z = 1$.
Thus, for $x = 3$, we have found only two values of $y$ for which condition $∃z(x=y*z)$ applies : the first one is $1$ and the second one is $3$, i.e. $x$ itself.
Thus the conditional is true for those values, while it is vacuously true for all other values of $y$ different from $1$ and $3$.
In conclusion, for $x = 3$, we have that $¬(x=1)∧∀y(∃z(x=y*z)⇒(y=x)∨(y=1))$ is true, i.e. $P(x)$ holds.
For $x = 4$, we can check that, with $y = 2$ we have $z = 2$ [i.e. : $4 = 2*2$]. For $y = 2$ we have that the antecedent of the conditional is satisfied but the consequent is not ($2$ is different from both $1$ and $4$). Thus, we have found a factor of $x = 4$ which is not $1$ or $x$ itself.
In this case, it is not true that $∃z(x=y*z)⇒(y=x)∨(y=1)$ holds for any $y$, and we conclude that $P(x)$ does not hold for $4$.
Conclusion : the correct answer is (A) $P(x)$ being true means that $x$ is a prime number.
Your approach is correct. Suppose that $\forall xP(x) \vee \forall xQ(x)$ is true and $\forall x(P(x) \vee Q(x))$ is false. Negating the quantifier and applying De Morgan's law gives $\exists x(\neg P(x) \wedge \neg Q(x))$, so that $P(a)$ and $Q(a)$ are false for some $a$ in the domain. However, two applications of universal instantiation imply that either $P(a)$ or $Q(a)$ are true. This is a contradiction, since $P(a)$ and $Q(a)$ are false.
An easy way to check check formulas of propositional and predicate logic for validity is the method of semantic tableaux, which gives a proof procedure for predicate (and other more exotic) logics.
Best Answer
domain of discourse: days of $2021$
$F(x)$ := $\;x$ is Friday
domain of discourse: $\mathbb Z$
$F(x)$ := $\;x+1=x.$