Rule of Distribution

Definition
Conjunction distributes over disjunction:

$$ $$

Disjunction distributes over conjunction:

$$ $$

Its abbreviation in a tableau proof is $$\text{Dist}$$.

Alternative rendition
These can alternatively be rendered as:

$$ $$ $$ $$

They can be seen to be logically equivalent to the forms above.

Proof by Natural Deduction
By the Tableau method:

Then we use the Rule of Commutation:

Proof by Truth Table
We apply the Method of Truth Tables to the proposition.

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

$$\begin{array}{|ccccc||ccccccc|} \hline p & \land & (q & \lor & r) & (p & \land & q) & \lor & (p & \land & r) \\ \hline F & F & F & F & F & F & F & F & F & F & F & F \\ F & F & F & T & T & F & F & F & F & F & F & T \\ F & F & T & T & F & F & F & T & F & F & F & F \\ F & F & T & T & T & F & F & T & F & F & F & T \\ T & F & F & F & F & T & F & F & F & T & F & F \\ T & T & F & T & T & T & F & F & T & T & T & T \\ T & T & T & T & F & T & T & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$$

$$\left({q \lor r}\right) \land p \dashv \vdash \left({q \land p}\right) \lor \left({r \land p}\right)$$ is demonstrated similarly.

Proof by Natural Deduction
By the Tableau method:

Then we use the Rule of Commutation:

Proof by Truth Table
We apply the Method of Truth Tables to the proposition.

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

$$\begin{array}{|ccccc||ccccccc|} \hline p & \lor & (q & \land & r) & (p & \lor & q) & \land & (p & \lor & r) \\ \hline F & F & F & F & F & F & F & F & F & F & F & F \\ F & F & F & F & T & F & F & F & F & F & T & T \\ F & F & T & F & F & F & T & T & F & F & F & F \\ F & T & T & T & T & F & T & T & T & F & T & T \\ T & T & F & F & F & T & T & F & T & T & T & F \\ T & T & F & F & T & T & T & F & T & T & T & T \\ T & T & T & F & F & T & T & T & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$$

$$\left({q \land r}\right) \lor p \dashv \vdash \left({q \lor p}\right) \land \left({r \lor p}\right)$$ is demonstrated similarly.