Proof by Cases/Formulation 1/Forward Implication/Proof 1
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Theorem
- $\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {p \implies r} \land \paren {q \implies r}$ | Premise | (None) | ||
| 2 | 1 | $p \implies r$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 3 | 1 | $q \implies r$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
| 4 | 1 | $p \lor q \implies r \lor r$ | Sequent Introduction | 2, 3 | Constructive Dilemma | |
| 5 | 5 | $p \lor q$ | Assumption | (None) | ||
| 6 | 1, 5 | $r \lor r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 4, 5 | ||
| 7 | 1, 5 | $r$ | Sequent Introduction | 6 | Rule of Idempotence: Disjunction | |
| 8 | 1 | $p \lor q \implies r$ | Rule of Implication: $\implies \II$ | 5 – 7 | Assumption 5 has been discharged |
$\blacksquare$