Proving DeMorgan's Law in IPC with Weak Law of Excluded Middle I am trying to show that $\neg (A \land B)$ implies $\neg A \lor \neg B$ in Intuitionistic Propositional Calculus (IPC) with the added inference rule of ''Weak excluded middle," which is that you can invoke $\neg \neg P \lor \neg P$.
My idea is to assume assume $\neg \neg A \lor \neg A$. If $\neg A$, the result follows. So we have to deal with $\neg \neg A$. So assume either $\neg \neg B \lor \neg B$. If $\neg B$, the result follows. So we have $\neg \neg A \land \neg \neg B$. Assumedly, one can prove $\neg (\neg A \lor \neg B)$ from this, and then finish the proof. But I am stuck on that last step of showing  $\neg \neg A \land \neg \neg B \Rightarrow \neg (\neg A \lor \neg B)$. I know this follows from the truth tables, but I am trying to do this just using the inference rules of the system I'm working in.
 A: Welcome to MSE!
You have exactly the right idea.
We want to show that, under weak LEM, $\lnot(A \land B)$ implies $\lnot A \lor \lnot B$. How do we do this? Well, just as you've said, by weak LEM we may assume $\lnot A \lor \lnot \lnot A$ and $\lnot B \lor \lnot \lnot B$.
Since the $\lnot A$ and $\lnot B$ cases both immediately give the claim, we're left to consider the case of $\lnot \lnot A \land \lnot \lnot B$. It's also worth remembering that we haven't used our $\lnot(A \land B)$ assumption yet!
Of course, it's well known that $\lnot \lnot A \land \lnot \lnot B$ is equivalent to $\lnot \lnot (A \land B)$, and from $\lnot(A \land B)$ and $\lnot \lnot (A \land B)$ we can prove false, and from there, the claim.
As an aside, weak LEM is actually intuitionistically equivalent to $\lnot (A \land B) \to \lnot A \lor \lnot B$, and now that you've seen one direction, you might try to go the other way too!

I hope this helps ^_^
A: In case anyone wants to see how this is done without recourse to a Heyting algebra (as I was thinking, when I first asked it), one can use as a lemma that double negation distributes over implication intuitionistically (Does double negation distribute over implication intuitionistically?), and then perform the following (at 14 we are using the lemma, but hopefully elsewhere what move we are making is clear) 
