Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
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
2  
@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
12  
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
add comment

4 Answers 4

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.
share|improve this answer
add comment

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

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
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
show 1 more comment

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.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.