Absorption Laws (Logic)/Conjunction Absorbs Disjunction

From ProofWiki
Jump to navigation Jump to search

Theorem

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


This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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


Proof 1

We apply the Method of Truth Tables.

As can be seen by inspection, the appropriate truth values match for all boolean interpretations.

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

$\blacksquare$


Proof 2

By calculation:

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

$\blacksquare$


Sources