I know how to prove tautologies with truth tables, but not with semantic definitions. How can I prove, using the semantic definitions, that the following are tautologies?
\begin{gather} \bigl(\exists x\,P(x) \lor \exists x\,Q(x)\bigr) \to \exists x\,\bigl(P(x) \lor Q(x)\bigr) \tag{1} \\ \bigl(\forall x\,P(x) \land \forall x\,Q(x)\bigr) \to \forall x\,\bigl(P(x) \land Q(x)) \tag{2} \end{gather}
In some circles, the term tautology is used to mean just those sentences that are true based on their truth functional connectives. Sentences in first-order logic that must be true based on the semantics of the logic other than truth-functional connectives are often called the validities of the language.
In any case, we're interested in semantic arguments for the truth of sentences. In the propositional case, we might ask for a semantic justification of the formula $P \to (P \lor Q)$. To demonstrate this with truth tables is straightforward, if somewhat tedious. To take the semantic approach, however, we look at the connectives in the sentence and what they mean.
$(P \land R)\to (P \lor Q)$ is a conditional, and a conditional is true when the truth of the antecedent implies the truth of the consequent.
So, does the truth of the antecedent imply the truth of the consequent? The truth of the antecedent requires that $P$ be true, which means that at least one of $P$ and $Q$ is true, so the consequent must also be true. Therefore, the conditional is true.
This type of reasoning carries over to the first-order case, but we must handle a few more types of sentences. Particularly, a first-order interpretation fixes a non-empty domain of discourse and maps each constant symbol to an element of the domain, each $n$-ary relation symbol to a $n$-ary relation on the domain. An interpretation $\cal I$ satisfies an expression $P(a)$ if and only if the set that the interpretation maps $P$ to contains the individual that the interpretation maps $a$ to, i.e., if $\cal I(a) \in \cal I(P)$. $\cal I$ satisfies a universal quantification $\forall x.\phi(x)$ if for it satisfies $\phi(x)$ for every $x$ in the domain, and $\exists x.\phi(x)$ if it satisfies $\phi(x)$ for some $x$ in the domain.
Let's consider the sentence $\forall x.(P(x) \lor \lnot P(x))$. This is a first-order validity. Why? Well, it is true if for each $x$ in the domain, $P(x) \lor \lnot P(x)$ is true. This is true if either $P(x)$ is true or $\lnot P(x)$ is true. $P(x)$ is true if $x$ is in $\cal I(P)$. $\lnot P(x)$ is true if $P(x)$ is false. $P(x)$ is false if $x \not\in \cal I(P)$. Since $x$ must be in $\cal I(P)$ or not in $\cal I(P)$, at least one of these cases must be true, and so the sentence is true.
As another example, consider $\forall x.P(x) \to \exists x.P(x)$. This is true if the truth of $\forall x.P(x)$ implies the truth of $\exists x.P(x)$. Without going through all the low level details, $\forall x.P(x)$ is true if every $x$ in the domain is in $\cal I(P)$. Since the domain is non-empty, that means that there is at least one $x$ such that $x \in \cal I(P)$, and that is enough to make $\exists x.P(x)$ true. Therefore, the truth of the antecedent implies the truth of the consequent, so the conditional is true.
These kinds of explanations can be rather tedious in some ways, especially because they sometimes seem so obvious. We may be tempted to say, “well of course $P \land Q$ being true implies that $P$ is true and $Q$ is true; that's what $\land$ means!” But that's the whole point—logic systems have two sides: a syntactic side and a semantic side, and we have to be able to see the connection between the two. We tend to be pretty good at this when the syntax has symbols the meanings of which we understand, but with more complicated languages, it is important to be able to do this rigorously. Other times, we might be interested in exploring different possible semantics for languages. For instance, in standard first-order logic, an interpretation is based on a non-empty domain of discourse. If we consider a different semantics that allows empty domains of discourse, then semantic analysis will show that $\forall x.(P(x) \lor \lnot P(x))$ is a validity under both kinds of semantics, but $\exists x.(P(x) \lor \lnot P(x))$ is a validity only in the semantics that requires a non-empty domain of discourse. This means that sentences like $\forall x.(P(x) \lor \lnot P(x)) \to \exists x.(P(x) \lor \lnot P(x))$ will be true under one semantics and not the other. This means that the same set of inference rules will be sound for one semantics, but unsound for the other. We can only discover this if we can reason about these sentences semantically.