Factor Principles/Conjunction on Left/Formulation 1/Proof 2
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Theorem
- $p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies q$ | Premise | (None) | ||
2 | 2 | $r \land p$ | Assumption | (None) | ||
3 | 1, 2 | $p$ | Rule of Simplification: $\land \EE_2$ | 2 | ||
4 | 1, 2 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
5 | 1, 2 | $r$ | Rule of Simplification: $\land \EE_1$ | 2 | ||
6 | 1, 2 | $r \land q$ | Rule of Conjunction: $\land \II$ | 5, 4 | ||
7 | 1 | $\paren {r \land p} \implies \paren {r \land q}$ | Rule of Implication: $\implies \II$ | 2 – 6 | Assumption 2 has been discharged |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Theorem $18$