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

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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

$\blacksquare$