How to prove that "$\forall x(P(x)\vee Q(x))$” and "$\forall xP(x)\vee\forall xQ(x)$” are not equivalent? How to prove that $”\forall x (P(x)\lor Q(x))”$ and $”\forall xP(x)\lor\forall xQ(x)”$ are not equivalent?
How to prove it? I don't even know how to start.
 A: When you're trying to prove two statements are not necessarily equivalent, one generally tries to find a specific circumstance in which the two statements are not equivalent.
When it comes to first-order logic with quantifiers, this means we need the following:

*

*A set $A$ that the quantifiers range over (depending on the book, this is sometimes required to be a nonempty set)


*For each predicate symbol, an "interpretation" of the predicate as an actual predicate on the set $A$


*For each function symbol, an "interpretation" of the function as an actual function on the set $A$
In this case, I will choose my set to be $\{0, 1\}$.
There are two predicate symbols which occur in your statements: $P$, and $Q$. I will interpret $P(x)$ to be the statement $x = 0$, and I will interpret $Q(x)$ to be the statement $x = 1$.
We now interpret the statement $\forall x (P(x) \lor Q(x))$ as the statement "For all $x \in \{0, 1\}$, $x = 0$ or $x = 1$." This is clearly a true statement.
We also interpret the statement $\forall x P(x)$ as "For all $x \in \{0, 1\}$, $x = 0$." This is a false statement, since $1 \in \{0, 1\}$ and $1 \neq 0$.
We also interpret the statement $\forall x Q(x)$ as "For all $x \in \{0, 1\}$, $x = 1$." This is a false statement, since $0 \in \{0, 1\}$ and $0 \neq 1$.
Thus, we see that the statement $\forall x P(x) \lor \forall x Q(x)$ is not true in this interpretation, since both $\forall x P(x)$ and $\forall x Q(x)$ are not true.
In this interpretation, $\forall x (P(x) \lor Q(x))$ is true, but $\forall x P(x) \lor \forall x Q(x)$ is false. Thus, the two statements are not logically equivalent.
