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I read the operation in the "An Introduction to Formal Logic" by Smith P.enter image description here It shows that in the QL "If everyoneX is such that then so" is truth-equivalent to “SomeoneX is such that if X is such that then so”, but it seems implausible in the case where "if everyone died in a war then human being would disappear " and "someone is such that if he died in a war then human being would disappear"

PS: It seems that even if we replaced “Fn” with “A” the operation was still established, was`nt it? enter image description here

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  • I don't see how "If everyone died in a war then human being would disappear" is a possible instantiation of the formulae Smith uses above. In Fn the predicate F has to be the same predicate as in ∀xFx. n seems to be intended as a constant. So you could have a case like: "If everyone died in a war, then Nathan died in a war." That'd be equivalent to "There's someone such that if they died in a war, Nathan died in a war." But you can't alter the meaning of Fn to involve a completely different predicate (like "would disappear") from the Fx bound by ∀x. Am I missing something? Feb 28, 2022 at 12:50
  • Check by cases: if Fn is True, both are true. Now consider the case when Fn is False. Feb 28, 2022 at 14:35
  • @AlabamaScholiast I get you but it seems that even if we replaced “Fn” with “A” the operation was still established. And I don`t think "the same predicate“ is of great significance for we might have a case: "If everyone got extinct in a war then human being would get extinct" . And in your giving case,it seems to be the domain of discourse that matters,if the x in ∀xFx included the n in Fn then the“∀xFx→Fn” would be tautology.
    – Pure
    Mar 1, 2022 at 7:35
  • @MauroALLEGRANZA Well, I can`t get what you mean by “true” and "false“. Do they refer to "tautologically true" and "tautologically false"?
    – Pure
    Mar 1, 2022 at 7:40
  • @Pure: I agree that if the domain of discourse includes n then ∀xFx → Fn is a tautology. (For example, if Ned is part of the relevant everybody, then "If everybody ate pizza, then Ned ate pizza" is tautologically true.) But then so is ∃x(Fx → Fn), under the same conditions: "There exists someone such that: if that person ate pizza, then Ned ate pizza." This is also tautologically true so long as n is in the domain of discourse. There must exist at least one x such that Fx → Fn, because even if no-one else, n is guaranteed to be that one. Mar 1, 2022 at 15:39

2 Answers 2

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In the example you quote, the equivalence holds because the conditional being used (⊃) is the material conditional. This is a special kind of conditional that is a truth function and is always true when its antecedent is false. It is highly useful in formal logic, but it is a long way from being a general account of the meaning of 'if' in English. So, while ∃x(Fx ⊃ Fn) is a logical truth of classical predicate logic, it should not be read as "there is some particular x, such that, if x is F then n is F". Indeed, more generally, whenever you see an existential quantifier sitting in front of a formula whose main connective is a material conditional, you should take great care when interpreting it.

The most natural understanding of the sentence, "there is someone such that if he died in a war then all human beings would disappear" is to take it as a counterfactual conditional, expressing what would be the case if something hypothetically were to happen. Conditionals that range over hypothetical possibilities like this are not material conditionals and require a different logical treatment.

This example is similar to the Drinker Paradox discussed by Raymond Smullyan, although unfortunately Smullyan himself did not seem to realise that the paradox can be explained as due to the limitations of the material conditional.

C S Pierce also discussed some examples similar to this and concluded that ordinary English conditionals are not truth functions.

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  • So, I agree that the interpretation of formulae containing material implication requires great care, and I also agree that it requires great care because of mismatches between intuitive meanings of natural language if-then conditions & the truth-functional meaning of the φ⊃ψ operator. (I also see how the use of the M.I. replacement rule in the last step of the passage above would highlight this issue.) Mar 2, 2022 at 15:20
  • However, I'm not sure how this really applies to understanding issue above. If Fn Then Fn is tautologically true on inspection, & (thus) so is ∃x(If Fx Then Fn) even with a really robustly non-truth-functional interpretation of the If, wherever n is a constant within the domain of discourse for our quantifiers. If it's not part of the domain of discourse, ∃x(If Fx Then Fn) wouldn't be true; but in that case neither would If: ∀xFx Then: Fn. Truth conditions will always be the same on either side of the ≡ whether conditional's interpreted as material, strict or any other way. Mar 2, 2022 at 15:29
  • If there is a problem here between logical formulae & NL meaning, it seems more like it'd be a problem over existential import not over material implication. Mar 2, 2022 at 15:31
  • Thanks for the comments. "∃x(If Fx Then Fn)" is not guaranteed to be true if it means something like: "there is some constructively identifiable value of x such that if x is F then n is F", since this does not assure us that this value of x can be identified with n. But "If ∀xFx Then Fn" is still good. One might argue that in this case, the mismatch is due to a stronger interpretation of the existential quantifier that blocks the instantiation rule.
    – Bumble
    Mar 2, 2022 at 19:57
  • But there are more complicated issues that I believe do involve the interpretation of the conditional. Suppose our domain is citizens of ancient Rome in 44 BC, one of whom is called Nigellus, and F is interpreted as murderer of Julius Caesar. We know that there is at least one such murderer, but not how many or who. "If ∀xFx Then Fn" holds, since if everybody murdered Caesar then Nigellus did, and if everybody had murdered Caesar this would have included Nigellus...
    – Bumble
    Mar 2, 2022 at 19:57
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  • Fx and Fn are predicates, so they must have the same meaning whether in the form bound to a quantifier or as instantiated with the constant n. For example, "If everyone ate pizza, then Nelly ate pizza" and "There is someone of whom the following is true: if they ate pizza, then Nelly ate pizza" is a possible instance of these formulae; similarly, so is "If everyone died in a war, then Nate died in a war" and "There is someone of whom the following is true: if that person died in a war, then Nate died in a war." (There is a problem in the original question, because you don't use the same predicate on both sides of the if-then, even though the formulae require the same predicate; e.g., "___ died in a war" vs. "___ would disappear." If you later edit the answer to fix this problem, anyone should feel free to edit this answer to remove this aside in the parentheses.)

  • n appears to be intended as a constant. Provided that this constant is within the domain of discourse, both of your formulae here are tautologically true.

    • You've already indicated in comments that you see that this is so on the left-hand side: ∀xFx → Fn must be true, simply in virtue of universal instantiation. "If everybody dreamed it would rain, then Nielsen dreamed it would rain." -- So long as Nielsen is part of the relevant domain of discourse when we talk about "everybody," then yes, of course.

    • But if so, then so is the right-hand side, under the same conditions: ∃x(Fx → Fn) must be true, simply in virtue of existential generalization. Wherever the constant n is part of the domain of discourse, Fn → Fn is tautologically true ("If Natasha lit the fuse, then Natasha lit the fuse"). But then there must exist at least one x such that Fx → Fn, because even if no-one else is, n is guaranteed to be that one. "There is someone of whom the following is true: if they lit the fuse, then Natasha lit the fuse." -- At least one, yes, because that is true of Natasha if no-one else.

  • One reason for seeing both of these as equivalent is because they are both tautologies, meaning that they both have the same truth-conditions in all cases (wherever n is part of the domain of discourse, both formulae are always true).

    • If you want to look at the steps of the transformation that Smith provides in the passage that you cite, the way that they get from one side of the equivalence to the other is by means of using the interdefinability of quantifiers -- ~∀xFx is equivalent to ∃x~Fx (not all x are F = some x is non-F) -- followed by an allowable expansion to the scope of the existential quantifier.

    • If you want to think about this in more general terms, you might consider that both the left-hand side and the right-hand side are fundamental consequences of the relationships between the quantifiers and the domain of discourse; provided that a constant n is within the domain of discourse, there are certain things that have to be true when quantifying over a domain of discourse that includes it, among them the fact that it's available as an instantiation for the universal quantifier, and also the fact that it's available as an instantiation for the existential quantifier.

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  • I know ∃x(Fx → Fn) and ∀xFx → Fn are both tautologically true if Fx and Fn are the same predicates,I mentioned " ∀xFx → Fn" is tautologically true just to show that if the Fx and Fn were the same predicates then the discourse about the truth of ∃x(Fx → Fn) and ∀xFx → Fn would be little realistic meaning.
    – Pure
    Mar 2, 2022 at 9:55
  • And first and foremost,the predicate doesn't matter at all,because we replaced “Fn” with “A” ,then the operation above was still established.
    – Pure
    Mar 2, 2022 at 9:57
  • If you change the formulae in the equivalence, then some of my remarks apply and others do not. In particular, Fn → A is not tautologically true in the way that Fn → Fn is, so my comments about existential generalization would not be applicable as it is to the original case. In that case, the relevant issue would most likely have much more to do with the semantics of material implication, which @Bumble discusses in their answer above. Do those remarks answer the question you had in mind? Mar 2, 2022 at 16:01
  • Under truth-functional MI rules, (i) suppose A to be true; then ∀xFx → A and ∃xFx → A are always true regardless of the truth of the antecedent; ∃x(Fx → A) is true provided there is at least one x in the domain of discourse. But then (ii) suppose B to be false. Then the conditional ∀xFx → A will be true iff there is at least one non-F; but then, if there is at least one non-F, then ∃x(Fx → A) will also be true because there's at least one x for which the antecedent Fx is false; material implications w false antecedents always turn out true, regardless of the consequent. Mar 2, 2022 at 16:09
  • This has the ugly consequence that material implication rules require treating seemingly counterintuitive conditionals as true, for example, If everyone sneezes, then the sun will rise in the east tomorrow (where antecedent & consequent appear unrelated but consequent is true), or There's someone of whom it's true that: if they are a unicorn, then I am emperor of the universe (where antecedent & consequent appear unrelated but antecedent is always false). Mar 2, 2022 at 16:19

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