Principle of Dilemma/Formulation 2/Forward Implication
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Theorem
- $\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$
Proof 1
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {p \implies q} \land \paren {\neg p \implies q}$ | Assumption | (None) | ||
| 2 | 1 | $q$ | Sequent Introduction | 1 | Principle of Dilemma: Formulation 1 | |
| 3 | $\paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$
Proof 2
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {p \implies q} \land \paren {\neg p \implies q}$ | Assumption | (None) | ||
| 2 | 1 | $p \implies q$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 3 | 1 | $\neg p \implies q$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
| 4 | 4 | $\neg q$ | Assumption | (None) | ||
| 5 | 1, 4 | $\neg p$ | Modus Tollendo Tollens (MTT) | 2, 4 | ||
| 6 | 1, 4 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 3, 5 | ||
| 7 | 1, 4 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 4, 6 | ||
| 8 | 1 | $\neg \neg q$ | Proof by Contradiction: $\neg \II$ | 4 – 7 | Assumption 4 has been discharged | |
| 9 | 1 | $q$ | Double Negation Elimination: $\neg \neg \EE$ | 8 | ||
| 10 | $\paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ | Rule of Implication: $\implies \II$ | 1 – 9 | Assumption 1 has been discharged |
$\blacksquare$