Can the material implication ever be used as the main connective within the scope of an existential quantifier?

Usually, a conjunction is the main connective in sentences bound by an existential quantifier. My text says this is

Because ‘(∃x)’ says something extremely minimal: ‘There is something'. So we usually need to build up the features of that something to say something interesting

'Usually' is not 'always'.

The wording doesn't preclude the main connective being something other than the conjunction. For example, if I wanted to notate:

  • There are apples or there are oranges. ∃x(Ax ∨ Ox)
  • There is something such that, if it is a nuke, it is also a WMD. ∃x(Nx → Wx)

So is it not necessarily the case that the main connective of a sentence bound with an existential quanitifer be a conjunction?

  • $\begingroup$ Sorry, I hit return too early on my previous version of this comment. Let me try again. $\endgroup$ – Jack Lee Dec 16 '13 at 23:29
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    $\begingroup$ Your second symbolic example doesn't say what the corresponding English sentence says. "There are some nukes only if there are some WMDs" would be symbolized as $(\exists x Nx) \rightarrow (\exists x Wx)$. Your symbolic statement actually means "There is something such that, if it is a nuke, it is also a WMD." $\endgroup$ – Jack Lee Dec 16 '13 at 23:32
  • $\begingroup$ Noted. I edited it. Funny, the notation got closer to what I meant than the words did. $\endgroup$ – Hal Dec 16 '13 at 23:38

There is an answer to this question, such that if the question is asking for a specific example, then the answer is providing it.

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    $\begingroup$ Ha! You're the coolest guy I've met today. Witty. $\endgroup$ – Hal Dec 17 '13 at 0:05
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    $\begingroup$ Thanks! I'm glad you like it... $\endgroup$ – JDH Dec 17 '13 at 0:08
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    $\begingroup$ That would be (∃x)((Aaq & Eq)→Pa) Axy: x is the answer to y; Ex: x is asking for a specific example; Px: providing a specific example; a = answer; q=question. How'd I do? $\endgroup$ – Hal Dec 17 '13 at 0:16
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    $\begingroup$ Looks good to me! $\endgroup$ – JDH Dec 17 '13 at 0:21

You're right that it makes most sense for the the topmost connective under an $\exists$ to be a conjunction (or an atomic formula).

Writing something else under the $\exists$ is not wrong, but it will rarely be the most natural to express the meaning you want.

For example $\exists$ distributes over disjunction, so instead of "there is something that is an apple or an orange" $\exists x.(Ax\lor Ox)$ it is usually more natural to think "there is an apple or there is an orange" $(\exists x.Ax)\lor(\exists x.Ox)$ or equivalently $(\exists x.Ax)\lor(\exists y.Oy)$ -- as can in fact be seen from the English phrasing you suggested for $\exists x.(Ax\lor Ox)$.

In classical logic $p\to q$ is the same as $\neg p\lor q$, so $\exists x.(Nx\to Wx)$ can be said simpler (?) as $(\exists x.\neg Nx)\lor(\exists y.Wy)$ "there is something that is not a nuke, or weapons of mass destruction exist". This makes it clearer that the actual contents of the statement doesn't really depend on whether any particular thing is simultaneously a nuke and a WMD.

Finally $\exists x.\neg p(x)$ is the same as $\neg\forall x.p(x)$. Here it is less clear which wording is the most natural. Sometimes it is most clear to say that there is something that isn't a pony; at other times it is more natural to say that not everything is a pony.

We can use the $\neg$ correspondence to rewrite $(\exists x.\neg Nx)\lor(\exists y.Wy)$ into $\neg(\forall x.Nx)\lor(\exists y.Wy)$, which we can then recognize as equivalent to an implication: $(\forall x.Nx)\to(\exists y.Wy)$. So

There is something such that, if it is a nuke, it is also a WMD.

is actually equivalent to

If all there is are nukes, then there is a WMD somewhere.

The latter formulation seems at least to me to give the more intuitive grasp of what statement claims.

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  • $\begingroup$ Thanks. Nice hat, by the way. $\endgroup$ – Hal Dec 17 '13 at 14:28

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