Principle of Composition/Formulation 1/Forward Implication
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Theorem
- $\paren {p \implies r} \lor \paren {q \implies r} \vdash \paren {p \land q} \implies r$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {p \implies r} \lor \paren {q \implies r}$ | Premise | (None) | ||
| 2 | 2 | $p \implies r$ | Assumption | (None) | ||
| 3 | 3 | $p \land q$ | Assumption | (None) | ||
| 4 | 3 | $p$ | Rule of Simplification: $\land \EE_ 1$ | 3 | ||
| 5 | 2, 3 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2 , 4 | ||
| 6 | 2 | $\paren {p \land q} \implies r$ | Rule of Implication: $\implies \II$ | 3 – 5 | Assumption 3 has been discharged | |
| 7 | 7 | $q \implies r$ | Assumption | (None) | ||
| 8 | 8 | $p \land q$ | Assumption | (None) | ||
| 9 | 8 | $q$ | Rule of Simplification: $\land \EE_ 2$ | 8 | ||
| 10 | 7, 8 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 7 , 9 | ||
| 11 | 7 | $\paren {p \land q} \implies r$ | Rule of Implication: $\implies \II$ | 8 – 10 | Assumption 8 has been discharged | |
| 12 | 1 | $\paren {p \land q} \implies r$ | Proof by Cases: $\text{PBC}$ | 1 , 2 – 6 , 7 – 11 | Assumptions 2 and 7 have been discharged |
$\blacksquare$