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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

  1. ∀x (Px → Qx)
  2. Px → Qx from 1, Universal instantiation
  3. ∃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)?

Lauro Morais
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  • Why do you think something's wrong with the proof? – Malice Vidrine Jul 14 '17 at 04:18
  • Because if it's ok, then universal quantifiers would suppose the existence of at least one member in the classes they range over, which would contradict (a). – Lauro Morais Jul 14 '17 at 04:30
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    It's usually taken as an assumption that the universe is non-empty. If the universe is empty, then it's silly: any universal sentence is true, and any existential false. We typically don't care about this border case. – Malice Vidrine Jul 14 '17 at 04:54
  • The "usual" semantics for classical logic is "an interpretation for the language is a structure $M = <d,I>$, where $d$ is a non-empty set, called the domain-of-discourse, or simply the domain, of the interpretation, and $I$ is an interpretation function." – Mauro ALLEGRANZA Jul 14 '17 at 07:20
  • This assumption licenses the inference: $\forall x \ Px \vDash \exists x \ Px$. – Mauro ALLEGRANZA Jul 14 '17 at 07:20

1 Answers1

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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 $x := x_0$.

goblin GONE
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  • Well then, the difference between the denotation of existential and universal quantifiers is only a matter of defining the domain? I mean the distinctions I made in (a) and (b) don't hold on non-empty domains? – Lauro Morais Jul 14 '17 at 04:41
  • @LauroMorais, I don't really understand the question, but the meaning of existential and universal quantifiers is completely different. If you think of $\mathrm{false}$ as $0$ and $\mathrm{true}$ as $1$, then $\exists$ can be interpreted as $\max$ and $\forall$ as $\min$. – goblin GONE Jul 14 '17 at 04:47
  • I use meaning in frege-russelian sense of the word. Sorry let me explain it better: universal quantifiers assure us the existence of classes (i.e., they have meaning), but they don't tell us if these classes have members (i.e., denotation), existential quanfieirs, by the other hand, assure us both the existence of classes and of at least a member of these classes. However, if an existential quantifier follows from a universal quantifiers it means this universal quantifier implied both the existence of the class and of a member of this class, just as the existential quantifier does. – Lauro Morais Jul 14 '17 at 05:04
  • In this case, universal quantifiers would have both meaning and denotation. – Lauro Morais Jul 14 '17 at 05:05