Proof by Cases/Formulation 1/Reverse Implication
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Theorem
- $\paren {p \lor q} \implies r \vdash \paren {p \implies r} \land \paren {q \implies r}$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \lor q} \implies r$ | Premise | (None) | ||
2 | 2 | $p$ | Assumption | (None) | ||
3 | 2 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 2 | ||
4 | 1, 2 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
5 | 1 | $p \implies r$ | Rule of Implication: $\implies \II$ | 2 – 4 | Assumption 2 has been discharged | |
6 | 6 | $q$ | Assumption | (None) | ||
7 | 6 | $p \lor q$ | Rule of Addition: $\lor \II_2$ | 6 | ||
8 | 1, 6 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 7 | ||
9 | 1 | $q \implies r$ | Rule of Implication: $\implies \II$ | 6 – 8 | Assumption 6 has been discharged | |
10 | 1 | $\paren {p \implies r} \land \paren {q \implies r}$ | Rule of Conjunction: $\land \II$ | 5, 9 |
$\blacksquare$