Proof by Cases with Contradiction
Theorem
- $\vdash p \iff \paren {p \lor q} \land \paren {p \lor \neg q}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p$ | Assumption | (None) | ||
| 2 | 1 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 1 | ||
| 3 | 1 | $p \lor \neg q$ | Rule of Addition: $\lor \II_2$ | 1 | ||
| 4 | 1 | $\paren {p \lor q} \land \paren {p \lor \neg q}$ | Rule of Conjunction: $\land \II$ | 2, 3 | ||
| 5 | $p \implies \paren {p \lor q} \land \paren {p \lor \neg q}$ | Rule of Implication: $\implies \II$ | 1 – 4 | Assumption 1 has been discharged | ||
| 6 | 6 | $\paren {p \lor q} \land \paren {p \lor \neg q}$ | Assumption | (None) | ||
| 7 | 6 | $p \lor \paren {q \land \neg q}$ | Sequent Introduction | 6 | Disjunction Distributes over Conjunction | |
| 8 | 8 | $p$ | Assumption | (None) | ||
| 9 | $\map \neg {q \land \neg q}$ | Theorem Introduction | (None) | Principle of Non-Contradiction: Formulation 2 | ||
| 10 | 6 | $p$ | Modus Tollendo Ponens $\mathrm {MTP}_2$ | 7, 9 | ||
| 11 | 6 | $p$ | Proof by Cases: $\text{PBC}$ | 6, 8 – 8, 9 – 10 | Assumptions 8 and 9 have been discharged | |
| 12 | $\paren {p \lor q} \land \paren {p \lor \neg q} \implies p$ | Rule of Implication: $\implies \II$ | 6 – 11 | Assumption 6 has been discharged | ||
| 13 | $p \iff \paren {p \lor q} \land \paren {p \lor \neg q}$ | Biconditional Introduction: $\iff \II$ | 5, 12 |
$\blacksquare$
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 $\text{T69}$