Rule of Simplification/Sequent Form/Formulation 1/Form 2
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Theorem
\(\ds p \land q\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds q\) | \(\) | \(\ds \) |
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \land q$ | Premise | (None) | ||
2 | 1 | $q$ | Rule of Simplification: $\land \EE_2$ | 1 |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
$\begin{array}{|ccc||c|} \hline p & \land & q & q \\ \hline \F & \F & \F & \F \\ \F & \F & \T & \T \\ \T & \F & \F & \F \\ \T & \T & \T & \T \\ \hline \end{array}$
As can be seen, when $p \land q$ is true so is $q$.
$\blacksquare$
Sources
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $3$: The Method of Deduction: $3.2$: The Rule of Replacement