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)?
Exists iff for all
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logic
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0Note 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. – 2011-06-19
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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. – 2011-06-19
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0What about $Z\neq\varnothing$? – 2011-06-19
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0@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. – 2011-06-19
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12Why 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. – 2011-06-19