Rule of Distribution

Context
Natural deduction

Definition
Conjunction distributes over disjunction:


 * $$p \land \left({q \lor r}\right) \vdash \left({p \land q}\right) \lor \left({p \land r}\right)$$
 * $$\left({p \land q}\right) \lor \left({p \land r}\right) \vdash p \land \left({q \lor r}\right)$$


 * $$\left({q \lor r}\right) \land p \vdash \left({q \land p}\right) \lor \left({r \land p}\right)$$
 * $$\left({q \land p}\right) \lor \left({r \land p}\right) \vdash \left({q \lor r}\right) \land p$$

Disjunction distributes over conjunction:


 * $$p \lor \left({q \land r}\right) \vdash \left({p \lor q}\right) \land \left({p \lor r}\right)$$
 * $$\left({p \lor q}\right) \land \left({p \lor r}\right) \vdash p \lor \left({q \land r}\right) $$


 * $$\left({q \land r}\right) \lor p \vdash \left({q \lor p}\right) \land \left({r \lor p}\right)$$
 * $$\left({q \lor p}\right) \land \left({r \lor p}\right) \vdash \left({q \land r}\right) \lor p$$

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

Proof: Conjunction distributes over Disjunction
These are proved by the tableau method:

$$p \land \left({q \lor r}\right) \vdash \left({p \land q}\right) \lor \left({p \land r}\right)$$:

$$\left({p \land q}\right) \lor \left({p \land r}\right) \vdash p \land \left({q \lor r}\right)$$

We can use the Rule of Commutation to show:

$$\left({q \lor r}\right) \land p \vdash \left({q \land p}\right) \lor \left({r \land p}\right)$$

$$\left({q \land p}\right) \lor \left({r \land p}\right) \vdash \left({q \lor r}\right) \land p$$

Proof: Disjunction distributes over Conjunction
$$p \lor \left({q \land r}\right) \vdash \left({p \lor q}\right) \land \left({p \lor r}\right)$$

$$\left({p \lor q}\right) \land \left({p \lor r}\right) \vdash p \lor \left({q \land r}\right) $$

We can use the Rule of Commutation to show:

$$\left({q \land r}\right) \lor p \vdash \left({q \lor p}\right) \land \left({r \lor p}\right)$$

$$\left({q \lor p}\right) \land \left({r \lor p}\right) \vdash \left({q \land r}\right) \lor p$$

Q.E.D.