Rule of Simplification/Sequent Form/Formulation 1/Form 1

Theorem

\(\ds p \land q\)

\(\)

\(\ds \)

\(\ds \vdash \ \ \)

\(\ds p\)

\(\)

\(\ds \)

Proof 1

By the tableau method of natural deduction:

$p \land q \vdash p$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \land q$	Premise	(None)
2		1	$p$	Rule of Simplification: $\land \EE_1$	1

$\blacksquare$

Proof by Truth Table

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

• 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables: Exercise $\text{I} \ \mathbf b.$
• 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $7.$