Proof by Contradiction/Variant 2/Formulation 2
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Theorem
- $\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \implies q} \land \paren {p \implies \neg q}$ | Assumption | (None) | ||
2 | 1 | $p \implies q$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
3 | 1 | $p \implies \neg q$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
4 | 1 | $\neg p$ | Sequent Introduction | 2, 3 | Proof by Contradiction: Formulation 1 | |
5 | $\paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$ | Rule of Implication: $\implies \II$ | 1 – 4 | 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': $\S 3$: Theorem $\text{T34}$