Rule of Material Implication/Formulation 1/Reverse Implication/Proof 2
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Theorem
- $\neg p \lor q \vdash p \implies q$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg p \lor q$ | Premise | (None) | ||
| 2 | 1 | $\neg \paren {\neg \neg p \land \neg q}$ | Sequent Introduction | 1 | De Morgan's Laws: Disjunction | |
| 3 | 1 | $\neg \neg p \implies q$ | Sequent Introduction | 2 | Conditional is Equivalent to Negation of Conjunction with Negative: $\neg \paren {p \land \neg q} \vdash p \implies q$ | |
| 4 | 4 | $p$ | Assumption | (None) | ||
| 5 | 4 | $\neg \neg p$ | Double Negation Introduction: $\neg \neg \II$ | 4 | ||
| 6 | 1, 4 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 3, 5 | ||
| 7 | 1 | $p \implies q$ | Rule of Implication: $\implies \II$ | 4 – 6 | Assumption 4 has been discharged |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Theorem $48$