Proof by Cases with Contradiction
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Theorem
- $\vdash p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$
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 | $\left({p \lor q}\right) \land \left({p \lor \neg q}\right)$ | Rule of Conjunction: $\land \II$ | 2, 3 | ||
5 | $p \implies \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$ | Rule of Implication: $\implies \II$ | 1 – 4 | Assumption 1 has been discharged | ||
6 | 6 | $\left({p \lor q}\right) \land \left({p \lor \neg q}\right)$ | Assumption | (None) | ||
7 | 6 | $p \lor \left({q \land \neg q}\right)$ | Sequent Introduction | 6 | Disjunction Distributes over Conjunction | |
8 | 8 | $p$ | Assumption | (None) | ||
9 | $\neg \left({q \land \neg q}\right)$ | 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 | $\left({p \lor q}\right) \land \left({p \lor \neg q}\right) \implies p$ | Rule of Implication: $\implies \II$ | 6 – 11 | Assumption 6 has been discharged | ||
13 | $p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$ | 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 $\text{T69}$