De Morgan's Laws (Logic)

Theorems

 * $$\neg p \or \neg q \dashv \vdash \neg \left({p \and q}\right)$$
 * $$\neg p \and \neg q \dashv \vdash \neg \left({p \or q}\right)$$
 * $$p \and q \dashv \vdash \neg \left({\neg p \or \neg q}\right)$$
 * $$p \or q \dashv \vdash \neg \left({\neg p \and \neg q}\right)$$

Their abbreviation in a tableau proof are collectively $$\textrm{DM}$$.

Proof by Natural Deduction
By the tableau method:

Proofs that use the Law of the Excluded Middle
The following results require the Law of the Excluded Middle to prove, and hence are not accepted by the school of intuitionist logic.

Comment
Note that this:


 * $$\neg p \and \neg q \dashv \vdash \neg \left({p \or q}\right)$$

can be proved in both directions without resorting to the LEM.

All the others:


 * $$\neg p \or \neg q \vdash \neg \left({p \and q}\right)$$
 * $$p \and q \vdash \neg \left({\neg p \or \neg q}\right)$$
 * $$p \or q \vdash \neg \left({\neg p \and \neg q}\right)$$

are not reversible in intuitionist logic.

Proof by Truth Table
We apply the Method of Truth Tables to the propositions in turn.

As can be seen by inspection, in all cases the truth values under the main connectives match for all models.

$$\begin{array}{|ccccc||cccc|} \hline \neg & p & \or & \neg & q & \neg & (p & \and & q) \\ \hline T & F & T & T & F & T & F & F & F \\ T & F & T & F & T & T & F & F & T \\ F & T & T & T & F & T & T & F & F \\ F & T & F & F & T & F & T & T & T \\ \hline \end{array}$$

$$\begin{array}{|cccccc||ccc|} \hline \neg & (\neg & p & \or & \neg & q) & p & \and & q \\ \hline F & T & F & T & T & F & F & F & F \\ F & T & F & T & F & T & F & F & T \\ F & F & T & T & T & F & T & F & F \\ T & F & T & F & F & T & T & T & T \\ \hline \end{array}$$

$$\begin{array}{|ccccc||cccc|} \hline \neg & p & \and & \neg & q & \neg & (p & \or & q) \\ \hline T & F & T & T & F & T & F & F & F \\ T & F & F & F & T & F & F & T & T \\ F & T & F & T & F & F & T & T & F \\ F & T & F & F & T & F & T & T & T \\ \hline \end{array}$$

$$\begin{array}{|cccccc||ccc|} \hline \neg & (\neg & p & \and & \neg & q) & p & \or & q \\ \hline F & T & F & T & T & F & F & F & F \\ T & T & F & F & F & T & F & T & T \\ T & F & T & F & T & F & T & T & F \\ T & F & T & F & F & T & T & T & T \\ \hline \end{array}$$