Rule of Idempotence/Conjunction/Formulation 1

Theorem

The conjunction operator is idempotent:

$p \dashv \vdash p \land p$

This can be expressed as two separate theorems:

$p \vdash p \land p$

$p \land p \vdash p$

As can be seen by inspection, the truth values under the main connectives match for each boolean interpretations.

$\begin{array} {|c||ccc|} \hline p & p & \land & p \\ \hline \T & \T & \T & \T \\ \F & \F & \F & \F \\ \hline \end{array}$

$\blacksquare$

When invoking this version of the Rule of Idempotence in a tableau proof, use the Idempotence template:

{{Idempotence|line|pool|statement|depends|type}}

where:

line is the number of the line on the tableau proof where Rule of Idempotence is to be invoked

pool is the pool of assumptions (comma-separated list)

statement is the statement of logic that is to be displayed in the Formula column, without the $ ... $ delimiters

depends is the line (or lines) of the tableau proof upon which this line directly depends

type is the type of Rule of Idempotence whose link will be displayed in the Notes column: in this instance Conjunction.

1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Exercise $1 \ \text{(b)}$
2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$