Factor Principles/Conjunction on Right/Formulation 1/Proof 1
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Theorem
- $p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies q$ | Premise | (None) | ||
2 | $r \implies r$ | Law of Identity | (None) | This is a theorem so depends on nothing | ||
3 | 1 | $\paren {p \implies q} \land \paren {r \implies r}$ | Rule of Conjunction: $\land \II$ | 1, 2 | ||
4 | 1 | $\paren {p \land r} \implies \paren {q \land r}$ | Sequent Introduction | 3 | Praeclarum Theorema |
$\blacksquare$