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

== Theorem ==
Let $\struct {S, \wedge, \vee, \preceq}$ be a uniquely complemented lattice.
Then the following are equivalent:
$(1):\quad \forall p, q \in S: \neg p \vee \neg q = \neg \paren {p \wedge q}$
$(2):\quad \forall p, q \in S: \neg p \wedge \neg q = \neg \paren {p \vee q}$
$(3):\quad \forall p, q \in S: p \preceq q \iff \neg q \preceq \neg p$
$(4):\quad \struct {S, \wedge, \vee, \preceq}$ is a distributive lattice.
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose:
:$\forall p, q \in S: \neg p \vee \neg q = \neg \paren {p \wedge q}$
Then applying this to $\neg p$ and $\neg q$:
:$\neg \neg p \vee \neg \neg q = \neg \paren {\neg p \wedge \neg q}$
By Complement of Complement in Uniquely Complemented Lattice:
:$\neg \neg p = p$ and $\neg \neg q = q$
Thus:
:$p \vee q = \neg \paren {\neg p \wedge \neg q}$
Taking complements of both sides:
:$\neg \paren {p \vee q} = \neg \neg \paren {\neg p \wedge \neg q}$
Again applying Complement of Complement in Uniquely Complemented Lattice:
:$\neg \paren {p \vee q} = \neg p \wedge \neg q$
=== $(2)$ 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 $(1)$:
:$\neg q \vee \neg p = \neg \paren {q \wedge p}$
So:
:$\neg q \preceq \neg p \iff \neg \paren {q \wedge p} = \neg p$
Taking the complements of both sides of the equation on the , and applying Complement of Complement in Uniquely Complemented Lattice:
:$\neg q \preceq \neg p \iff q \wedge p = p$
But the is equivalent to $p \preceq q$. 
Therefore:
:$\neg q \preceq \neg p \iff p \preceq q$
=== $(3)$ implies $(1)$ ===
Suppose that $p \preceq q \iff \neg q \preceq \neg p$
By the definition of join:
:$\neg p, \neg q \preceq \neg p \vee \neg q$
Thus:
:$\neg \paren {\neg p \vee \neg q} \preceq p, q$
By the definition of meet:
:$\neg \paren {\neg p \vee \neg q} \preceq p \wedge q$
Thus:
:$\neg \paren {p \wedge q} \preceq \neg \neg \paren {\neg p \vee \neg q}$
By Complement of Complement in Uniquely Complemented Lattice:
:$*\quad \neg \paren {p \wedge q} \preceq \neg p \vee \neg q$
Dually:
:$\neg x \wedge \neg y \preceq \neg \paren {x \vee y}$
Letting $x = \neg p$ and $y = \neg q$:
:$\neg \neg p \wedge \neg \neg q \preceq \neg \paren {\neg p \vee \neg q}$
By Complement of Complement in Uniquely Complemented Lattice:
:$p \wedge q \preceq \neg \paren {\neg p \vee \neg q}$
By the premise and Complement of Complement in Uniquely Complemented Lattice:
:$**\quad \neg p \vee \neg q \preceq \neg \paren {p \wedge q}$
By $*$ and $**$:
:$\quad \neg \paren {p \wedge q} = \neg p \vee \neg q$
=== $(1)$, $(2)$, and $(3)$ together imply $(4)$ ===
$b, c \preceq b \vee c$, so
:$a \wedge b \preceq a \wedge \paren {b \vee c}$
:$a \wedge c \preceq a \wedge \paren {b \vee c}$
By the definition of join:
:$\paren {a \wedge b} \vee \paren {a \wedge c} \preceq a \wedge \paren {b \vee c}$
== Sources ==
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Category:Uniquely Complemented Lattices
Category:Boolean Lattices
Category:Distributive Lattices