Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2
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Theorem
- $p \land \paren {p \lor q} \dashv \vdash p$
Proof
By calculation:
| \(\ds p \land \paren {p \lor q}\) | \(=\) | \(\ds \paren {p \lor \bot} \land \paren {p \lor q}\) | Disjunction with Contradiction | |||||||||||
| \(\ds \) | \(=\) | \(\ds p \lor \paren {\bot \land q}\) | Disjunction is Left Distributive over Conjunction | |||||||||||
| \(\ds \) | \(=\) | \(\ds p \lor \bot\) | Conjunction with Contradiction | |||||||||||
| \(\ds \) | \(=\) | \(\ds p\) | Disjunction with Contradiction |
$\blacksquare$