Rule of Conjunction/Sequent Form/Formulation 1/Proof by Truth Table
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Theorem
| \(\ds p\) | \(\) | \(\ds \) | ||||||||||||
| \(\ds q\) | \(\) | \(\ds \) | ||||||||||||
| \(\ds \vdash \ \ \) | \(\ds p \land q\) | \(\) | \(\ds \) |
Proof
We apply the Method of Truth Tables.
$\begin{array}{|c|c||ccc|} \hline p & q & p & \land & q\\ \hline \F & \F & \F & \F & \F \\ \F & \T & \F & \F & \T \\ \T & \F & \T & \F & \F \\ \T & \T & \T & \T & \T \\ \hline \end{array}$
As can be seen, only when both $p$ and $q$ are true, then so is $p \land q$.
$\blacksquare$