De Morgan's Laws imply Uniquely Complemented Lattice is Boolean Lattice

Theorem
Let $\left({S,\wedge,\vee,\preceq}\right)$ be a Definition:Uniquely Complemented Lattice.

Then the following are equivalent:

$(1)\quad \forall p, q \in S: \neg p \vee \neg q = \neg \left({p \wedge q}\right)$

$(2)\quad \forall p, q \in S: \neg p \wedge \neg q = \neg \left({p \vee q}\right)$

$(3)\quad \forall p, q \in S: p \preceq q \iff \neg q \preceq \neg p$

$(4)\quad \left({S,\wedge,\vee,\preceq}\right)$ is a Definition:Distributive Lattice.

$(1)$ implies $(2)$
Suppose:
 * $\forall p, q \in S: \neg p \vee \neg q = \neg \left({p \wedge q}\right)$

Let $x, y \in S$.

Then $\neg \neg x \vee \neg \neg y = \neg \left({\neg x \wedge \neg y}\right)$.

By Complement of Complement in Uniquely Complemented Lattice, $\neg \neg x = x$ and $\neg \neg y = y$.

Thus:
 * $x \vee y = \neg \left({\neg x \wedge \neg y}\right)$.

Taking complements of both sides:
 * $\neg \left({x \vee y}\right) = \neg \neg \left({\neg x \wedge \neg y}\right)$

Again applying Complement of Complement in Uniquely Complemented Lattice:


 * $\neg \left({x \vee y}\right) = \neg x \wedge \neg y$

$(2)$ implies $(1)$
By Dual Pairs (Order Theory), $\wedge$ and $\vee$ are dual.

Thus this implication follows from the above by Duality.

$(1)$ implies $(3)$
By the definition of a lattice:
 * $p \preceq q \iff p \vee q = q$

Applying this to $\neg q$ and $\neg p$:
 * $\neg q \preceq \neg p \iff \neg q \vee \neg p = \neg p$

By $(1)$:
 * $\neg q \vee \neg p = \neg \left({q \wedge p}\right)$

So:
 * $\neg q \preceq \neg p \iff \neg \left({q \wedge p}\right) = \neg p$

Taking the complements of both sides of the equation on the right, and applying Complement of Complement in Uniquely Complemented Lattice:
 * $\neg q \preceq \neg p \iff \left({q \wedge p}\right) = p$

But the right side is equivalent to $p \preceq q$

Therefore:
 * $\neg q \preceq \neg p \iff p \preceq q$