Set of Literals Satisfiable iff No Complementary Pairs

Theorem
Let $S$ be a set of literals.

Then $S$ is satisfiable it contains no complementary pairs.

Necessary Condition
Suppose $S$ contained the complementary pair $\set {p, \neg p}$.

By the Principle of Non-Contradiction, $\set {p, \neg p}$ is unsatisfiable.

By Superset of Unsatisfiable Set is Unsatisfiable, so is $S$.

Hence if $S$ is satisfiable, it cannot contain any complementary pairs.

Sufficient Condition
Suppose $S$ contains no complementary pair.

Define the boolean interpretation $v: \PP_0 \to \set {\T, \F}$ by:


 * $\map v p = \begin{cases}

\T & \text{if } p \in S \\ \F & \text{otherwise} \end{cases}$

Let $\mathbf A \in S$ be an arbitary literal.

If $\mathbf A = p$ is a positive literal, then $\map v {\mathbf A} = \T$ is true by the first clause of the definition.

Now suppose $\mathbf A = \neg p$ is a negative literal.

Then $p \notin S$, for $S$ contains no complementary pairs.

Therefore $\map v p = \F$, and hence:


 * $\map v {\neg p} = \map v {\mathbf A} = \T$

Since $v$ has been demonstrated to be a model of $S$, $S$ is satisfiable.