TITLE: Do De Morgan's laws hold in propositional intuitionistic logic?
QUESTION [31 upvotes]: In Wikipedia page on intuitionistic logic, it is stated that excluded middle and double negation elimination are not axioms. Does this mean that De Morgan's laws, stated $$ \lnot (p \land q) \iff \lnot p \lor \lnot q \\ \lnot (p \lor q) \iff \lnot p \land \lnot q,$$
cannot be proven in propositional intuitionistic logic?

REPLY [2 votes]: Here's a finite example that basically reproduces what's happening in Zhen Lin's answer/counterexample:

Take a 4-element boolean lattice (labelled 1-2-3-4 above) and a "supertop" to it (labelled 5, above). The elements of the antichain of this lattice (2 and 3) are still pseudocomplements of each other, although no longer complements, which goes hand-in-hand with the "enriched" lattice not being Boolean anymore. (You can check that e.g. $2\land x \le1$ [still] has the greatest solution 3.) It's also easy to see that this lattice is still distributive (because it has exactly 5 elements that's an easy check), so a Heyting lattice because it's finite and distributive.
Now, the meet of the two elements of the antichain is still the bottom element, but because we added the new "supertop" 5, the psedocomplement of the bottom (labelled 1) is going to be 5 in this enriched lattice (because $1 \land x \le 1$ now admits a greater solution (5) than 4 which was the top of the Boolean lattice we started with, so $\neg (2 \land 3) = 5$ now, whereas the "supertop" does nothing for the join of the preudocomplents of the elements of the antichain, i.e. $\neg 2 \lor \neg 3= 3 \lor 2 = 4$.
This is para is not a proof, but note that this supertop doesn't affect the other
De Mogran law: $\neg (2 \lor 3) = \neg 4 = 1$  and $\neg 2 \land \neg 3 = 3 \land 2 = 1$.
(Apologies for using 1 as the bottom element, I now this can be slightly confusing.)

The following part (also) isn't proof of anything, but it does give an additional intuition why only one of De Mogran's laws breaks in a Heyting algebra... and why I added a supertop above, instead of a superbottom, labelled 0 below:

Then, "nothing happens" as far De Morgan's laws are concerned because the definition of a pseudocomplement is relative to the (new) (super)bottom. So $\neg (2 \land 3) = \neg 1 = 0$ but $\neg 2 \lor \neg 3 = 0 \lor 0 = 0$ now as well because adding this superbottom (0) actually changed the pseudocomplements of both 2 and 3 (as well as that of 1).
Likewise this super-bottomed lattice fails to break the other De Morgan law: $\neg (2 \lor 3) = \neg 4 = 0$  and $\neg 2 \land \neg 3 = 0 \land 0 = 0$.