Rule of Simplification/Sequent Form/Formulation 1/Form 1/Proof by Truth Table
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Theorem
\(\ds p \land q\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds p\) | \(\) | \(\ds \) |
Proof
We apply the Method of Truth Tables.
$\begin{array}{|ccc||c|} \hline p & \land & q & p \\ \hline \F & \F & \F & \F \\ \F & \F & \T & \F \\ \T & \F & \F & \T \\ \T & \T & \T & \T \\ \hline \end{array}$
As can be seen, when $p \land q$ is true so is $p$.
$\blacksquare$
Sources
- 1988: Alan G. Hamilton: Logic for Mathematicians (2nd ed.) ... (previous) ... (next): $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Example $1.8 \ \text{(a)}$