Rule of Commutation/Conjunction/Formulation 1/Proof 1
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Theorem
- $p \land q \dashv \vdash q \land p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \land q$ | Premise | (None) | ||
2 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
3 | 1 | $q$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
4 | 1 | $q \land p$ | Rule of Conjunction: $\land \II$ | 3, 2 |
$\Box$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $q \land p$ | Premise | (None) | ||
2 | 1 | $q$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
2 | 1 | $p$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
4 | 1 | $p \land q$ | Rule of Conjunction: $\land \II$ | 3, 2 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Theorem $17$