TITLE: Triple negation in intuitionistic logic?
QUESTION [5 upvotes]: I'm told that intuitionistic logic is basically classical logic with "law of the excluded middle removed", and that you can't from $\neg \neg A$ infer $A$.
So, if we consider $A$ a state, $\neg A$ a state, and $\neg \neg A$ a state, does this mean that $\neg \neg \neg A$ is a separable state?

REPLY [2 votes]: Triple negation $\neg \neg \neg P $ is a shorthand for $((P \rightarrow \bot) \rightarrow \bot) \rightarrow \bot$. It turns out however that for any pair of propositions $P$ and $Q$, we have that $((P \rightarrow Q) \rightarrow Q) \rightarrow Q$ simplifies to $P \rightarrow Q$, which lets us simplify triple negation to $P \rightarrow \bot$ which is just the definition of $\neg P$.
To prove this, just note that by modus ponens, we have $P \rightarrow ( (P \rightarrow Q) \rightarrow Q)$, for any P, Q. But we note that the RHS of that is equal to the LHS of $((P \rightarrow Q) \rightarrow Q) \rightarrow Q$ , so we can just compose the two implications to get $P \rightarrow Q$. This holds for any logic with modus ponens and transitive implication.