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I have been presented with this syllogism:

All labradors are four legged (1)
All dogs are labradors (2)
Therefore, some dogs are four legged (3)

I responded valid, my reasoning as follows:

L = Labrador
F = Four Legged
D = Dog

∀x L(x) ->F(x) -- 1
∀x D(x) -> L(x) --2

The hypothesis:

∃x D(x) ∧ F(x)

I think I am allowed to begin my assumption by saying that if x is a dog, then by combining rules 1 and 2 then x is also four legged:

∀x (D(x) -> L(x) ∧ L(x) -> F(x)) -> ∀x ( D(x) -> F(x) ) -- 1+2

But - isn't the "is valid for all" statement trivially imply "is valid for some"?

Let me explain. If we have a generic predicate P, wouldn't it make sense that:

(a) ∀x P(x) |= ∃x P(x)

If that wasn't the case, then we'd have:

(b) ∀x P(x) |= ¬∃x P(x)

(Quite simply, because there either is or isn't an x that makes a given predicate valid, saying (c) ¬∃x P(x) ∧ ∃x P(x) would be a contradiction! )

Well, b must be false, as it translates to "P(x) being valid for all x entails that there is no x for which P(x) is valid" which is a contradiction. So by c we must accept a which allows me to (trivially) say that if all dogs are four legged, then some dogs are four legged.

..Any thoughts?

MrD
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  • You are right : $\forall x P(x) \vDash \exists x P(x)$ is correct in classical logic : if it holds for all, it follows that it holds for some. This principle is not true if we consider also "empty" universe of discourse ... but this is another issue. Aristotle's Syllogism has "existential import". You can see this post. – Mauro ALLEGRANZA Sep 10 '15 at 18:57
  • @MauroALLEGRANZA - Notice that allowing the empty domain does not make you a non-classical logician. Hodges and Poizat allow for the empty domain, but that does not make them non-classical (Poizat justifies this in terms of his back-and-forth method). – Nagase Sep 10 '15 at 19:03

1 Answers1

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Assuming a non-empty domain, it is indeed the case that $\forall x P(x) \models \exists x P(x)$. However, from that you can't conclude that $\forall x (P(x) \to Q(x)) \models \exists x (P(x) \wedge Q(x)$! In particular, if you have a model in which $\forall x \neg P(x)$, then $\forall x (P(x) \to Q(x))$ will be true (since nothing satisfies the antecedent), but $\exists x (P(x) \wedge Q(x))$ is false (since there's no $x$ which satisfies $P(x)$), e.g., take the domain to be $\{0\}$ and $P = \varnothing$.

So the syllogism isn't valid, as it could be the case that nothing is a dog, which would make the premises true, but the conclusion false.

Nagase
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