# 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.

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## 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:

- $\neg \paren {p \vee q} = \neg p \wedge \neg q$

$\Box$

### $(2)$ implies $(1)$

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

Thus this implication follows from the above by Duality.

$\Box$

### $(1)$ 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 right hand side, and applying Complement of Complement in Uniquely Complemented Lattice:

- $\neg q \preceq \neg p \iff q \wedge p = p$

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

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Therefore:

- $\neg q \preceq \neg p \iff p \preceq q$

$\Box$

### $(3)$ implies $(1)$

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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 $**$:

- $\quad \neg \paren {p \wedge q} = \neg p \vee \neg q$

$\Box$

### $(1)$, $(2)$, 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:

- $\paren {a \wedge b} \vee \paren {a \wedge c} \preceq a \wedge \paren {b \vee c}$

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## Sources

*This article incorporates material from Uniquely Complemented Lattice on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.*