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Theorem

$q \implies (q \lor p)$

Proof

By the definitional abbreviation for the conditional:

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

the Rule of Addition can be written as:

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

This evaluates as follows:

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

$\blacksquare$