There exists a superintuitionistic propositional logic where $\neg p \vee \neg \neg p$ is a theorem, but $p \vee \neg p$ is not a theorem. It is called the logic of the weak excluded middle. That raises the question, is there a superintuitionistic propositional logic where $\neg \neg p \vee \neg \neg \neg p$ is a theorem, but $\neg p \vee \neg \neg p$ is not a theorem? Of course, the question can be generalized by adding more negation signs.
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1Noah Schweber's answer aside, you could maybe ask questions about what if you include one of the higher entries of the Rieger-Nishimura lattice (the free Heyting algebra on one generator) as an axiom. So for example, ask about a logic where you assume $\lnot\lnot p \lor (\lnot\lnot p \rightarrow p)$; or about a logic where you assume $(\lnot\lnot p \rightarrow p) \rightarrow p \lor \lnot p$. – Daniel Schepler Sep 28 '20 at 22:00
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No - perhaps surprisingly, intuitionistic logic does prove triple negation elimination $$\neg\neg\neg p\equiv \neg p.$$ So this collapses the "LEM-hierarchy" right at the second level.
Noah Schweber
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