# Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2

## Theorem

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

## Proof

By calculation:

 $\ds p \land \left({p \lor q}\right)$ $=$ $\ds \left({p \lor \bot}\right) \land \left({p \lor q}\right)$ Disjunction with Contradiction $\ds$ $=$ $\ds p \lor \left({\bot \land q}\right)$ Disjunction is Left Distributive over Conjunction $\ds$ $=$ $\ds p \lor \bot$ Conjunction with Contradiction $\ds$ $=$ $\ds p$ Disjunction with Contradiction

$\blacksquare$