# Rule of Commutation/Disjunction/Formulation 2

## Theorem

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

### Forward Implication

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

## Proof 1

By the tableau method of natural deduction:

$\vdash \paren {p \lor q} \iff \paren {q \lor p}$
Line Pool Formula Rule Depends upon Notes
1 1 $p \lor q$ Assumption (None)
2 1 $q \lor p$ Sequent Introduction 1 Disjunction is Commutative
3 $\paren {p \lor q} \implies \paren {q \lor p}$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
4 4 $q \lor p$ Assumption (None)
5 4 $p \lor q$ Sequent Introduction 4 Disjunction is Commutative
6 $\paren {q \lor p} \implies \paren {p \lor q}$ Rule of Implication: $\implies \II$ 4 – 5 Assumption 4 has been discharged
7 $\paren {p \lor q} \iff \paren {q \lor p}$ Biconditional Introduction: $\iff \II$ 3, 6

$\blacksquare$

## Proof by Truth Table

We apply the Method of Truth Tables.

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

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

$\blacksquare$