Rule of Simplification/Sequent Form/Formulation 1

Theorem

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

\(\ds p \land q\)

\(\)

\(\ds \)

\(\ds \vdash \ \ \)

\(\ds p\)

\(\)

\(\ds \)

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

\(\ds p \land q\)

\(\)

\(\ds \)

\(\ds \vdash \ \ \)

\(\ds q\)

\(\)

\(\ds \)

Form 1

\(\ds p \land q\)

\(\)

\(\ds \)

\(\ds \vdash \ \ \)

\(\ds p\)

\(\)

\(\ds \)

Form 2

\(\ds p \land q\)

\(\)

\(\ds \)

\(\ds \vdash \ \ \)

\(\ds q\)

\(\)

\(\ds \)

Proof

We apply the Method of Truth Tables.

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

As can be seen, when $p \land q$ is true so are both $p$ and $q$.

$\blacksquare$