# Definition:Constructed Semantics/Instance 5/Rule of Commutation

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## Theorem

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

## Proof

By the definitional abbreviation for the conditional:

$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$

the Rule of Commutation can be written as:

$\neg \left({p \lor q}\right) \lor \left({q \lor p}\right)$

This evaluates as follows:

$\begin{array}{|cccc|c|ccc|} \hline \neg & (p & \lor & q) & \lor & (q & \lor & p) \\ \hline 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 2 & 0 & 2 & 0 & 0 \\ 1 & 0 & 0 & 3 & 0 & 3 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 3 & 1 & 2 & 2 & 0 & 2 & 2 & 1 \\ 0 & 1 & 3 & 3 & 0 & 3 & 3 & 1 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 & 2 \\ 3 & 2 & 2 & 1 & 0 & 1 & 2 & 2 \\ 3 & 2 & 2 & 2 & 0 & 2 & 2 & 2 \\ 1 & 2 & 0 & 3 & 0 & 3 & 0 & 2 \\ 1 & 3 & 0 & 0 & 0 & 0 & 0 & 3 \\ 0 & 3 & 3 & 1 & 0 & 1 & 3 & 3 \\ 1 & 3 & 0 & 2 & 0 & 2 & 0 & 3 \\ 0 & 3 & 3 & 3 & 0 & 3 & 3 & 3 \\ \hline \end{array}$

$\blacksquare$