# Is $\;\exists x.\lnot p\left(x\right) \rightarrow \exists x. p\left(x\right),\;$ and vice versa, always true?

I have a feeling this could be true for all cases, for example when I state that some fruits are not apples, does not this automatically mean that some fruits are (and vice versa)? That is, is it true that:

$$\exists x.\neg p\left(x\right) \to \exists x. p\left(x\right)\quad?$$

And what about: $$\;\exists x. p(x) \rightarrow \exists x.\lnot p(x)\quad?$$

On the other hand, I could not come up with any formal proof so I would like to hear your thoughts as to why I'm right or wrong about this.

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Just curious, what intuition makes you think this is true? – Thomas Andrews Jan 3 at 20:12
@ThomasAndrews cause if this was true it would solve my homework :) – Anton Jan 3 at 20:19
If you see a pig that cannot fly, you want to conclude that some pigs can fly? – Hagen von Eitzen Jan 3 at 20:21
@HagenvonEitzen Of course not, I was completely wrong. – Anton Jan 3 at 20:23

$$\exists x.\lnot p\left(x\right) \not\rightarrow \exists x. \,\,p\left(x\right)\tag{1}$$

$$\exists x.\;\;p(x) \not\rightarrow \exists x. \lnot p(x)\tag{2}$$

$(1)$ Within a given domain (I'll use human beings in my counter-example to your proposed implications), the existence of someone without a property does not imply the existence of someone with the property.

There exists humans who are not bears. There do not exist humans who are bears.

Suppose it's true that all humans who exist are not bears; we can still assert (truthfully) that therefore, there exist humans who are not bears. But it does not follow that there must therefore exist humans who are not bears.

$(2)$ Within a given domain (again, I'll use human beings in a counterexample to your vice versa "claim"), the existence of some $x$ such that $p(x)$ is true does not imply the existence of an $x$ such that it is not the case that $p(x)$ holds.

There exist humans who sleep. But there do not exist humans who don't sleep.

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That is up for debate. :-) – Asaf Karagila Jan 3 at 20:11

Nope. "There exist some oranges that are not apples" is a perfectly true statement, whereas "there exist some oranges that are apples" is perfectly false.

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Definitely not! Let $p(x)$ be any assertion which is always false, such as $\lnot(x=x)$.

And the vice-versa part is not correct either, for basically the same reason.

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Neither is always true. Some fruits are not apples would still be true even if no fruits were apples. For a mathematical example, let $p(x)$ mean ‘$x$ is an integer, and $x$ is both odd and even’. Then $\exists x\big(\neg p(x)\big)$ is certainly true, since $\neg p(1)$ is true. However, $\exists x\big(p(x)\big)$ is certainly not true, since $\forall x\big(\neg p(x)\big)$ is true.

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Say that $p(x)$ means "$x$ is an invisible pink unicorn". Your claim says that if there is an $x$ such that $x$ is not an invisible pink unicorn, then there is an $x$ such that $x$ is an invisible pink unicorn.

My left boot is not an invisible pink unicorn. Therefore, if your formula were correct, the existence of my boot would be a proof of the existence of invisible pink unicorns.

If your formula were correct, my boot would also be proof of the existence of omnipotent gods, of space aliens, of holes to the center of the Earth, and of flying spaghetti monsters, since it isn't any of those things either.

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