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I have two questions, which are probably easy explained but I am quite confused.

1- If I write: ∀xDx → ∃xAx, then the x in ∀xDx and the x in ∃xAx can be different things in an interpretation? Should they mean the same thing?

2- Is ∀yDy → ∃xAx = ∀y∃x(Dy → Ax) ?If it is, then what happens with ∀xDx → ∃xAx = ∀x∃x(Dx → Ax)? Wouldn't it be confusing?

Thanks.

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  • $\begingroup$ 1) $\forall x Dx$ means "every thing is $D$". $\exists x Ax$ means "some thing is $A$". Thus, they do not "mean the same thing". $\endgroup$ Dec 14, 2017 at 13:36
  • $\begingroup$ My question with the first one is whether the x could be a black cat and the x could be a Siamese dog (assume there are Siamese dogs). I understand that the same thing can have different predicates, my question is whether the same variable can be different "things" in the same sentence. $\endgroup$ Dec 14, 2017 at 13:46
  • $\begingroup$ Correct me if I am wrong, but then in question 2 the first example is correct and the second is not? Or neither of them are correct? $\endgroup$ Dec 14, 2017 at 13:48

1 Answer 1

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The equivalences with nested quntifiers are a little bit tricky...

In general, we have that: $(\alpha \to \exists x \ \beta) \equiv \exists x \ (\alpha \to \beta)$, provided that $x$ is not free in $\alpha$.

Thus, $(∀yDy \to ∃xAx) \equiv ∃x(∀yDy \to Ax)$, because $x$ is not free in $∀yDy$.

Nothing change if we write $∀xDx$.

But (alas!) we have that: $(\forall x \alpha \to \beta) \equiv \exists x \ (\alpha \to \beta)$, provided that $x$ is not free in $\beta$.

So, if we apply the rule to the formula above, what we get is:

$(∀yDy \to ∃xAx) \equiv ∃x(∀yDy \to Ax) \equiv ∃x∃y(Dy \to Ax)$.


Finally, we have that: $(∀xDx \to ∃xAx) \equiv ∃x(Dx \to Ax)$.

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  • $\begingroup$ Wow, any place where I can read further about this? My introductory book does not take such depth. $\endgroup$ Dec 14, 2017 at 13:59
  • $\begingroup$ @César Dalí Vázquez Navarro - Many textbooks... Kleene, Mathematical Logic or Simpson, Mathematical Logic. $\endgroup$ Dec 14, 2017 at 14:05
  • $\begingroup$ Grazie mille :) $\endgroup$ Dec 14, 2017 at 14:07

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