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Imagine a scenario where it can be proven that $\forall x \in U\; P(x),$ but only so by contradiction.

That is, assuming $\exists x \in U \; \neg P(x)$ leads to a contradiction, therefore we are forced to conclude $\forall x \in U\; P(x).$

But we just don't have any constructive, or even known, example of an $x \in U$ for which $P(x)$ holds.

"Non-Constructive Case" in the sense that we, till now, don't even now how to construct such an $x.$

Would be great to have an example of mathematical scenario where this happens ! and where, preferably, $U$ is not the empty set, for that would be just trivial vacuous truth.

medium_o
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  • But the inference from $\exists x \lnot$ to a contradiction to $\forall x$is contructively valid. Wht is not is from $\forall x \lnot$ to a contradiction to $\exists x$. – Mauro ALLEGRANZA Sep 11 '22 at 06:44
  • @MauroALLEGRANZA I think I used the word "constructive" in sense of "constructing an example/concrete case." Basically like "Oh! we have $\forall x ; P(x),$ but we don't know any example of such an $x$ – medium_o Sep 11 '22 at 06:54
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    But if we know that "everything is P", every object in the domain of the interpretation will be an example. – Mauro ALLEGRANZA Sep 11 '22 at 08:04
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    This question doesn't really have anything to do with $P(x)$; if you've proven $\forall x \in U, P(x)$ then, by definition, any element $x \in U$ satisfies $P(x)$. So your question is just "is there an example of a set $U$ that we know is non-empty but such that we can't construct any elements of it"? For that see: https://math.stackexchange.com/questions/575835/are-there-any-non-constructive-proofs-for-which-an-example-was-never-constructed – Qiaochu Yuan Sep 11 '22 at 08:14
  • @QiaochuYuan I'd say that's exactly what the question boils down to. Great ! – medium_o Sep 11 '22 at 10:20

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