Constructive Dilemma/Formulation 3
Theorem
- $\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} }$ | Assumption | (None) | ||
| 2 | 1 | $\paren {p \implies q} \land \paren {r \implies s}$ | Rule of Simplification: $\land \EE_2$ | 1 | Cutting corners: should use Associativity first | |
| 3 | 1 | $\paren {p \lor r} \implies \paren {q \lor s}$ | Sequent Introduction | 2 | Constructive Dilemma: Formulation 1 | |
| 4 | 1 | $p \lor r$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 5 | 1 | $q \lor s$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 3, 4 | ||
| 6 | $\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$ | Rule of Implication: $\implies \II$ | 1 – 5 | Assumption 1 has been discharged |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T48}$