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This is probably a very basic question but I haven't been able to make progress on it.

In intuitionistic logic, the law of excluded middle is rejected. This means that there may be propositions which are neither true nor false: $$\exists p : \neg (p \lor \neg p).$$ Using De Morgan's laws, this statement can be rewritten as $$\exists p : (\neg p) \land (\neg \neg p).$$ However, this seems to violate the law of noncontradiction because $\neg p$ and $\neg \neg p$ can't both be true.

It seems to me that in intuitionistic logic, De Morgan's law that $\neg (p \lor q) \to (\neg p) \land (\neg q)$ can only be true for the special case in which $q \not\to \neg p$ or $\neg p \not\to q$ (so that $\neg p \leftrightarrow q$ does not hold) but it cannot be true in general. But I don't see this mentioned anywhere so I must be making a mistake.

Where is the gap in my understanding?

Edit: Answers to the question Do De Morgan's laws hold in propositional intuitionistic logic? do not answer my question. The highest rated answer under that question states that $\neg p \land \neg q \dashv \vdash \neg (p \lor q)$, that is, $\neg (p \lor q) \leftrightarrow (\neg p) \land (\neg q)$. My question is about a contradiction that I seem to get when I apply this formula by substituting $\neg p$ for $q$.

Student
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  • Thanks @TankutBeygu, I have edited my question because the question you have linked does not answer my question. – Student Apr 01 '25 at 18:48
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    @MauroALLEGRANZA both directions of the equivalence $\neg (p \lor q) \leftrightarrow \neg p \land \neg q$ are valid intuitionistically. It is the other one of de Morgan's laws that only holds in one direction, and the incorrect reasoning in this post is due to assuming the version of that law for $\forall$/$\exists$ instead of $\land$/$\lor$ (see my answer). – Naïm Camille Favier Apr 01 '25 at 19:48
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    Why was this question closed? It is not a duplicate AFAICT. @amWhy – Naïm Camille Favier Apr 02 '25 at 00:48

1 Answers1

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In intuitionistic logic, the law of excluded middle is rejected. This means that there may be propositions which are neither true nor false

This is where you go wrong. Intuitionistic logic is simply classical logic without the law of excluded middle; it cannot prove that there are propositions that are neither true nor false, since classical logic does not prove that. In fact intuitionistic logic proves the negation of that fact: see the beginning of Andrej Bauer's Five stages of accepting constructive mathematics.

If you go further than intuitionistic logic and postulate the negation of the law of excluded middle, $\neg(\forall p. p \lor \neg p)$, you are still not able to conclude that $\exists p. \neg (p \lor \neg p)$, because the implication $\neg(\forall x. P(x)) \to \exists x. \neg P(x)$ is not valid intuitionistically.

Note that your comment about de Morgan's laws is incorrect: intuitionistically, the equivalence $\neg(p \lor q) \leftrightarrow \neg p \land \neg q$ is always valid, as is the implication $\neg p \lor \neg q \to \neg (p \land q)$; the converse implication $\neg (p \land q) \to \neg p \lor \neg q$ is not. A similar statement can be made for the quantifiers: the equivalence $\neg (\exists x. P(x)) \leftrightarrow \forall x. \neg P(x)$ is always valid, as is the implication $\exists x. \neg P(x) \to \neg \forall x. P(x)$, but the converse is not. This is why your reasoning fails.

Naïm Camille Favier
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