Proof by Contradiction/Variant 3/Formulation 2/Proof 1
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Theorem
- $\vdash \paren {p \implies \neg p} \implies \neg p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies \neg p$ | Assumption | (None) | ||
2 | 1 | $\neg p$ | Sequent Introduction | 1 | Proof by Contradiction: Variant 3: Formulation 1 | |
3 | $\paren {p \implies \neg p} \implies \neg p$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$