De Morgan's Laws (Boolean Algebras)

Theorem
Let $\left({S, \vee, \wedge, \neg}\right)$ be a Boolean algebra.

Then for all $a, b \in S$:


 * $\neg \left({a \vee b}\right) = \neg a \wedge \neg b$
 * $\neg \left({a \wedge b}\right) = \neg a \vee \neg b$

Proof
By virtue of Complement in Boolean Algebra is Unique, it will suffice to verify:


 * $\left({a \vee b}\right) \wedge \left({\neg a \wedge \neg b}\right) = \bot$
 * $\left({a \vee b}\right) \vee \left({\neg a \wedge \neg b}\right) = \top$

For the first of these, compute:

By the Duality Principle, we also conclude:


 * $\left({a \wedge b}\right) \vee \left({\neg a \vee \neg b}\right) = \top$

Substituting $\neg a$ and $\neg b$ for $a$ and $b$, respectively, this becomes:


 * $\left({\neg a \wedge \neg b}\right) \vee \left({a \vee b}\right) = \top$

where Complement of Complement (Boolean Algebras) was applied.

Since $\vee$ is commutative, we have also derived the second of the two equations above.

Hence:


 * $\neg \left({a \vee b}\right) = \neg a \wedge \neg b$

Now by again the Duality Principle:


 * $\neg \left({a \wedge b}\right) = \neg a \vee \neg b$

follows immediately.

Also see

 * De Morgan's Laws (Logic)