I have some doubts which I divided into two parts. Suppose the following universal sentence ∀x (Px → Qx) (i) Could I use universal generalization on it? I mean
- ∀x (Px → Qx)
- Px → Qx from 1, Universal instantiation
- ∃x (Px → Qx) from 2, Existential generalization
What's wrong with this proof? Don't the universal quantifiers actually suppose, or imply, existential quantifiers? (ii) What's the actual difference between universal and existential quantifiers? From what I gathered (a) universal quantifiers have meaning, but not necessarily denotation (i.e., they express classes but don't they assure these have members), whereas (b) existential quantifiers necessarily have both meaning and denotation (i.e., the classes these propositions express have at least one member). If universal quantifiers don't necessarily have denotation, would this mean that they only have meaning too (i.e., universal quantifiers have only meaning and no denotation at all)? In any cases, if the proof above holds, wouldn't it imply that the classes which universal quantifiers range over have at least one member and so contradict (a)?
Some implementations of first-order logic support the inference $$\forall xPx \rightarrow \exists x Px,$$ some don't. In terms of semantics, the question is whether or not we allow the empty domain.
If we do allow the empty domain, then the above inference fails. For let $P : \emptyset \rightarrow \{\mathrm{true},\mathrm{false}\}$ denote the unique such function. Then $\forall x \in \emptyset,P(x)$ is true, but $\exists x \in \emptyset,P(x)$ is false.
If we don't allow the empty domain, then the above inference holds. For let $X$ denote an arbitrary non-empty set, and let $P : X \rightarrow \{\mathrm{true},\mathrm{false}\}$ denote any such function. Assume $$\forall x \in X,P(x).$$ Then since $X$ is non-empty, we can find $x_0 \in X$. Hence by the above formula, we have $P(x_0)$. From this we deduce $\exists x \in X,P(x)$, namely $x := x_0$.