In intuitionistic logic, de Morgan’s law often refers to the one of de Morgan's four laws that is not an intuitionistic tautology, namely $\neg (P \wedge Q) \to (\neg P \vee \neg Q)$ for any $P,Q$.

Theorem

De Morgan’s law is equivalent to weak excluded middle.

Proof

If de Morgan’s law holds, then since $\neg (P \wedge \neg P)$, we have $\neg P \vee \neg\neg P$, as desired. Conversely, if weak excluded middle holds and we have $\neg (P\wedge Q)$, then from weak excluded middle we get $\neg P \vee \neg\neg P$ and $\neg Q \vee \neg\neg Q$ which give four cases. In three of those cases $\neg P \vee \neg Q$ holds, while in the fourth we have $\neg\neg P$ and $\neg\neg Q$, which together imply $\neg\neg(P\wedge Q)$ (see the first lemma here and its proof), contradicting the assumption of $\neg (P\wedge Q)$; so the fourth case is impossible.