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It's a logical axiom that $\forall x(\alpha \to \beta) \to (\forall x \alpha \to \forall x \beta)$. However, it's generally not true that $\forall x(\alpha \to \beta) \leftarrow (\forall x \alpha \to \forall x \beta)$, except for some trivial case, say, $x$ doesn't occur free in either $\alpha$ or $\beta$.

What's the general condition that $\forall x(\alpha \to \beta) \leftrightarrow (\forall x \alpha \to \forall x \beta)$ holds? What about $\exists x(\alpha \to \beta) \leftrightarrow (\exists x \alpha \to \exists x \beta)$ ?

Metta World Peace
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1 Answers1

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I don't have any proof, but here is my intuition:

I doubt there is any general condition besides obvious

$$\forall x(\alpha \to \beta) \leftarrow (\forall x \alpha \to \forall x \beta). \tag{1}$$

The problem is that it might be true for many different reasons, e.g. the universe being of size one or perhaps the $\forall x \beta$ is true and it makes the implication trivial. Observe that $\forall x(\alpha \to \beta)$ means that a single $x$ satisfying $\alpha$ is enough to make $\beta$ true (some clarification: it does not mean that $\beta$ is true in general, it is true only for that particular $x$). On the other hand $\forall x \alpha \to \forall x \beta$ means that you need $\alpha$ to hold for all the $x$es to satisfy $\beta$. For example, let's assume the universe has more than one element, then:

$$\forall x \Big(P(x) \to \exists y \big(P(x) \land P(y)\big)\Big)$$ would be of first kind (because you can set $y = x$), while $$\forall x P(x) \to \forall x \exists y(P(x)\land P(y) \land x \neq y)$$

is the second kind, because you need to know, that there are at least two $x$es that satisfy $P$. So the general condition would be something like "from the $\forall x \alpha$ you need only one $x$ to make $\beta$ true", but that is exactly $(1)$.

I hope it explained something ;-)

dtldarek
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  • To the observation that a single $x$ satisfying $\alpha$ makes $\beta$ true, note that nowhere (in the general case) it said that $x$ is not a free variable of $\beta$. – Asaf Karagila Mar 08 '13 at 09:25
  • @AsafKaragila I don't understand, in my first example $x \in FV(\beta)$. I am aware that this is not the same as $\exists x\alpha \to \forall x \beta$, do you think I should make it more clear? – dtldarek Mar 08 '13 at 09:38
  • I am referring to this, "Observe that $\forall x(\alpha\rightarrow\beta)$ means that a single $x$ satisfying $\alpha$ is enough to make $\beta$ true." If $x$ is a free variable of $\beta$ then it doesn't meant that $\beta$ is true, just that some $x$ satisfies it. If for some other $x$, $\alpha$ is false it could just as well be that $\beta$ is not satisfied by that particular $x$. – Asaf Karagila Mar 08 '13 at 09:40
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    @AsafKaragila Oh, I see. Well, I thought it was obvious, but since it is not, I added some clarification. Thank you for pointing that out! – dtldarek Mar 08 '13 at 09:49