# Exists iff for all

I have a theorem of the following scheme: $Q \Leftrightarrow \exists x\in Z: P(x) \Leftrightarrow \forall x\in Z: P(x)$. How to simplify it (not to write $P(x)$ twice)?

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Note that $\forall\implies\exists$, but it is not usually the case otherwise. These theorems usually witness some very strong property, that if someone has then everyone has. – Asaf Karagila Jun 19 '11 at 19:04
@Asaf Karagila: $\forall\implies\exists$ may not hold when $Z$ is empty. I need to specifically say something which non-emptyness of $Z$ follows from. – porton Jun 19 '11 at 19:06
What about $Z\neq\varnothing$? – Asaf Karagila Jun 19 '11 at 19:16
@Asaf Karagila: $Z\neq\varnothing$ but I need to present some statement from which follows $Z\neq\varnothing$ as a theorem in a non-cumbersome way. To do it in a non-cumbersome way is the crux of my question. – porton Jun 19 '11 at 19:19
Why do you want to avoid writing $P(x)$ twice? Redundance is not (always) a sin and some results craftily stated with the minimal number of words may be difficult to understand some months later. – Did Jun 19 '11 at 19:22

I think most people would write something like the following:

Theorem. The following are equivalent:

1. $Q$;

2. There exists $x \in Z$ such that $P(x)$;

3. For every $x \in Z$, $P(x)$.

By the same token, I think most readers would find this easiest to read (because they are used to seeing such statements), rather than having to decode some clever logical statement.

If $P(x)$ stands for a long and complicated condition that you don't want to write out twice, I'd introduce some auxilary definition or notation. (Also, that way readers won't have to check that 2 and 3 really contain the same condition.)

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$$Z\neq\emptyset\land\{x\in Z\mid P(x)\}\in\{\emptyset,Z\}$$

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Nice. And with only one $Z$? (Just kidding...) – Did Jun 20 '11 at 13:02
This statement is always true when $Z$ is empty, but $(\exists x\in Z\ P(x))\iff(\forall x\in Z\ P(x))$ is always false when $Z$ is empty. – JDH Jun 20 '11 at 13:08
@JDH: Thanks, I fixed the answer accordingly. – joriki Jul 6 '11 at 11:00

Your question is ambiguous as you have written it, since it isn't clear if you mean to say that $Q$ asserts the biconditional, or that all three statements are equivalent, or that $Q$ is your theorem and it is asserting the equivalence of the other two clauses. These are distinct assertions. Many mathematicians write $P\iff Q\iff R$, when what they mean is $(P\iff Q)\wedge(Q\iff R)$. The waters are muddied further by the question of whether $Z$ is nonempty and whether you want the assertion to imply that or not.

If you mean to mean to say $[Q\iff (\exists x\in Z\ P(x))]$ and $[Q\iff \forall x\in Z\ P(x)]$, then my suggestion would be:

• $Q$ holds when some, or equivalently every, element of $Z$ has property $P$.

But what you wrote could also be interpreted as $Q\iff[(\exists x\in Z\ P(x))\iff (\forall x\in Z\ P(x))]$, in which case my suggestion would be:

• $Q$ holds when every element of $Z$ has property $P$, if any does.

That formulation presumes $Z$ is known to be non-empty in advance. If you want this to be part of the assertion (part of $Q$?), then it would seem that you want:

• $Q$ holds when $Z$ is nonempty, and every element of $Z$ has property $P$, if any does.

Finally, perhaps you mean just that $Q$ is your theorem, and it asserts the equivalence of the other clauses. (Note that $Q$ does not appear in Joriki and user6312's answers, who evidently interpreted your question this way.) In this case, of course, you don't need to mention $Q$. So if your theorem simply is $[\exists x\in Z\ P(x)]\iff [\forall x\in Z\ P(x)]$, then I suggest:

• $Z$ is nonempty and every element of $Z$ has property $P$, if any does.
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Using semi-formal notation, one could write

$$(\forall x, y \in Z) (P(x) \longrightarrow P(y))$$

There are more formal (but less readable) ways of writing the same thing.

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But wasn't it required not to write $P(x)$ twice? – t.b. Jun 19 '11 at 21:42
@Theo Buehler: Yes, that was the stated goal. But I wanted to mention the "standard" way one writes such things when doing logic. To me, the main problem was the somehow jarring combination of $\exists$ and $\forall$. – André Nicolas Jun 19 '11 at 22:19
I am not clear on how this expression avoids the $\exists$, as the OP said the existential quantifier was there to claim that $Z$ was non-empty. – Asaf Karagila Jun 20 '11 at 6:33
@Asaf Karagila: I now see from a comment addressed to you that the author wants to build non-emptiness of $Z$ into the definition. What I would then suggest is to separate out that assertion and not have it be part of an iff. – André Nicolas Jun 20 '11 at 7:54
@user6312: Yes, that was my suggestion as well, though not as put clearly. It seems this suggestion was not welcomed. – Asaf Karagila Jun 20 '11 at 11:05