Rule of Commutation

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Definition

This rule is two-fold:

$p \land q \dashv \vdash q \land p$
$p \lor q \dashv \vdash q \lor p$

Its abbreviation in a tableau proof is $\text{Comm}$.


Alternative rendition

These can alternatively be rendered as:

  • $\vdash \left({p \land q}\right) \iff \left({q \land p}\right)$
  • $\vdash \left({p \lor q}\right) \iff \left({q \lor p}\right)$


They can be seen to be logically equivalent to the forms above.


Proof

Proof by Natural Deduction

By the tableau method:

  • $p \land q \vdash q \land p$:
Line Pool Formula Rule Depends upon
1 1 $p \land q$ P (None)
2 1 $p$ $\land \mathcal E_1$ 1
3 1 $q$ $\land \mathcal E_2$ 1
4 1 $q \land p$ $\land \mathcal I$ 3, 2

$\blacksquare$


By the same technique we can show $q \land p \vdash p \land q$.


  • $p \lor q \vdash q \lor p$:
Line Pool Formula Rule Depends upon
1 1 $p \lor q$ P (None)
2 1 $p$ A (None)
3 2 $q \lor p$ $\lor \mathcal I_2$ 2
4 1 $q$ A (None)
5 4 $q \lor p$ $\lor \mathcal I_1$ 4
6 1 $q \lor p$ $\lor \mathcal E$ 1, 2-3, 4-5

$\blacksquare$


By the same technique we can show $q \lor p \vdash p \lor q$.




Proof by Truth Table

We apply the Method of Truth Tables to the propositions in turn.

As can be seen by inspection, in both cases, the truth values under the main connectives match for all models.


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

$\blacksquare$


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

$\blacksquare$


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