Sorry for the title, I could not find a better way to express it. So, I am preparing for the exam and we use, at school, Kenneth Rosen's Discrete Math book. There is an example in the book as the following:
Show that $\forall x (P(x) \land Q(x))$ is equivalent to $\forall x P(x) \land \forall x Q(x)$ where the same domain is used throughout.
And the book, just shows if one is True the other is. But, it does not count in the second one the following: $P(a) \land Q(b)$.
I am a little confused, because in the second one, the variables may be assigned with different values. So, if I cannot express it in the first one correctly since there is only 1 variable, how could they be equivalent.
I am sure there is an explanation, but I do not find a way to correctly search on this since my knowledge is still very limited.
Proof in the book:
To show that these statements are logically equivalent, we must show that they always take the same truth value, no matter what the predicates $P$ and $Q$ are, and no matter which domain of discourse is used. Suppose we have particular predicates $P$ and $Q$, with a common domain. We can show that $\forall x (P(x) \land Q(x))$ and $\forall x P(x) \land \forall Q(x)$ are logically equivalent by doing two things. First, we show that if $\forall x (P(x) \land Q(x))$ is true, then $\forall x P(x) \land \forall Q(x)$ is true. Second, we show that if $\forall x P(x) \land \forall Q(x)$ is true, then $\forall x (P(x) \land Q(x))$ is true.
So, suppose that $\forall x (P(x) \land Q(x))$ is true. This means that if a is in the domain, then $P(a) \land Q(a)$ is true. Hence, $P(a)$ is true and $Q(a)$ is true. Because $P(a)$ is true and $Q(a)$ is true for every element in the domain, we can conclude that $\forall x P(x)$ and $\forall x Q(x)$ are both true. This means that $\forall x P(x) \land \forall Q(x)$ is true.
Next, suppose that $\forall x P(x) \land \forall Q(x)$ is true. It follows that $\forall x P(x)$ is true and $\forall Q(x)$ is true. Hence, if a is in the domain, then $P(a)$ is true and $Q(a)$ is true [because $P(x)$ and $Q(X)$ are both true for all elements in the domain, there is no conflict using the same value of a here]. It follows that for all a, $P(a) \land Q(a)$ is true. It follows that $\forall x (P(x) \land Q(x))$ is true. We can now conclude that
$\forall x (P(x) \land Q(x)) ≡ \forall x P(x) \land \forall Q(x)$.
I feel like there is a misconception about how rules of logical inference involving universal quantifiers function.
In the formal theory of logical inferences there are two major rules concerning universal quatifiers: Universal Specificaion (U.S.) and Universal Generalization (U.G.). The former tells us that if something is true for every object then it's true for some arbitrary object from the domain, formally: $\forall x P(x)\rightarrow P(x)$, where $x$ is some arbitrary object. Instead of $x$ we can use any symbol we want which refers to some object in the domain of discourse. U.S. tells us that if a statement is true for some arbitrary object $x$ then it's true for every object in the domain, $P(x)\rightarrow \forall x P(x)$. The key here is that objects considered are arbitrary objects.
Yes, they can, but this variable assignment won't help us to proof that $\forall x (P(x) \land Q(x))$. The beauty of Universal Specificaion is that we can pick whatever element we want from the domain and still we know that the predicate will be true. In the given situation we pick the same element from the domain, namely $a$, for both universal statements. So for this element $P(a)$ holds and $Q(a)$ holds, therefore $P(a)\land Q(a)$ holds as well. Since the elements were chosen arbitrary we can apply the Universal Generalization and conclude that $\forall x (P(x) \land Q(x))$ holds.
Sometimes when doing proofs involving universal quantifiers choosing the right elements from the domain as the replacements for the variables during the application of U.S. is the most critical part of the proof.