Negation of Conditional implies Antecedent
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Theorem
- $\vdash \neg \paren {p \implies q} \implies p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\neg \paren {p \implies q}$ | Assumption | (None) | ||
2 | 1 | $p \land \neg q$ | Sequent Introduction | 1 | Conjunction with Negative is Equivalent to Negation of Conditional | |
3 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 2 | ||
4 | $\neg \paren {p \implies q} \implies p$ | Rule of Implication: $\implies \II$ | 1 – 3 | Assumption 1 has been discharged |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T21}$