Proof by Cases/Formulation 3
Jump to navigation
Jump to search
Theorem
- $\vdash \paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies r$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$ | Theorem Introduction | (None) | Constructive Dilemma: Formulation 3 | ||
2 | $\paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies \paren {r \lor r}$ | Substitution | 1 | $r \ / \ q$, $q \ / \ r$, $s \ / \ r$ | ||
3 | 3 | $\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r}$ | Assumption | (None) | ||
4 | 3 | $r \lor r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 | ||
5 | 3 | $r$ | Sequent Introduction | 4 | Rule of Idempotence: Disjunction | |
6 | $\paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies r$ | Rule of Implication: $\implies \II$ | 3 – 5 | Assumption 3 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 $\text{T49}$