Suppose you have a bounded, complemented lattice $\mathfrak{L} = \left<L, \vee, \wedge, \neg, 1, 0\right>$ that satisfies De Morgan's laws. I want to prove that this is an ortholattice. The first condition $a \leq b \implies \neg b \leq \neg a$ was straightforward, but I am struggling a lot with the second condition.
That is, given $\neg (a \vee b) = \neg a \wedge \neg b$ and $\neg (a \wedge b) = \neg a \vee \neg b$ for all $a, b \in L$, I'd like to show that
$$\neg\neg a = a$$ for all $a, b, \in L$.
When I looked this up in a textbook, what I saw was: $\neg \neg a = \neg (\neg a \vee \neg a) = \neg \neg a \wedge \neg \neg a = a \wedge a = a $
I can't, however, understand how one gets $\neg \neg a \wedge \neg \neg a = a \wedge a$ without assuming $\neg\neg a = a$.
I also tried proving this in some other way but failed.
I need one of two things:
An alternative proof, or
An explanation why $\neg \neg a \wedge \neg \neg a = a \wedge a$ doesn't assume $\neg\neg a = a$
This isn't true in general unless the lattice is uniquely complemented. For instance, consider the diamond lattice $M_3$, with elements $\{0,a,b,c,1\}$ such that $0 < a,b,c < 1$ and none of $a,b,c$ are comparable to each other. We can assign complements $\neg a = b$ and $\neg b = c$ and $\neg c = a$. You can check that de Morgan's laws hold, as does the order-reversing condition, but $\neg \neg a \neq a$.