Absorption Laws (Logic)/Conjunction Absorbs Disjunction

Theorem

$p \land \paren {p \lor q} \dashv \vdash p$


This can be expressed as two separate theorems:

Forward Implication

$p \land \paren {p \lor q} \vdash p$

Reverse Implication

$p \vdash p \land \paren {p \lor q}$


Proof by Truth Table

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:

\(\ds p \land \paren {p \lor q}\) \(=\) \(\ds \paren {p \lor \bot} \land \paren {p \lor q}\) 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

$\blacksquare$


Sources

  • 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorem $31$
  • 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.10$: Exercise $2.5$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.