Rule of Addition/Sequent Form/Formulation 2/Proof by Truth Table

Theorem

\(\text {(1)}: \quad\)

\(\ds \vdash p\)

\(\implies\)

\(\ds \paren {p \lor q}\)

\(\text {(2)}: \quad\)

\(\ds \vdash q\)

\(\implies\)

\(\ds \paren {p \lor q}\)

Proof

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives are $T$ for all boolean interpretations.

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

$\blacksquare$