Rule of Conjunction/Sequent Form/Formulation 1/Proof 1
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Theorem
\(\ds p\) | \(\) | \(\ds \) | ||||||||||||
\(\ds q\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds p \land q\) | \(\) | \(\ds \) |
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p$ | Premise | (None) | ||
2 | 2 | $q$ | Premise | (None) | ||
3 | 1, 2 | $p \land q$ | Rule of Conjunction: $\land \II$ | 1, 2 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Theorem $12$