Contradictory Consequent/Proof 1
Jump to navigation
Jump to search
Theorem
- $p \implies \bot \dashv \vdash \neg p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies \bot$ | Premise | (None) | ||
2 | 2 | $p$ | Assumption | (None) | ||
3 | 1, 2 | $\bot$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 2 | ||
4 | 1 | $\neg p$ | Proof by Contradiction: $\neg \II$ | 2 – 3 | Assumption 2 has been discharged |
$\Box$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\neg p$ | Premise | (None) | ||
2 | 2 | $p$ | Assumption | (None) | ||
3 | 1,2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 1, 2 | ||
4 | 1 | $p \implies \bot$ | Rule of Implication: $\implies \II$ | 2 – 3 | Assumption 2 has been discharged |
$\blacksquare$