Modus Ponendo Ponens/Sequent Form/Proof 2
Jump to navigation
Jump to search
Theorem
| \(\ds p\) | \(\implies\) | \(\ds q\) | ||||||||||||
| \(\ds p\) | \(\) | \(\ds \) | ||||||||||||
| \(\ds \vdash \ \ \) | \(\ds q\) | \(\) | \(\ds \) |
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies q$ | Premise | (None) | ||
| 2 | 2 | $p$ | Premise | (None) | ||
| 3 | 3 | $p$ | Assumption | (None) | ||
| 4 | 1, 3 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
| 5 | 5 | $\lnot p$ | Assumption | (None) | ||
| 6 | 2, 5 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 2, 5 | ||
| 7 | 2, 5 | $q$ | Rule of Explosion: $\bot \EE$ | 6 | ||
| 8 | $p \lor \lnot p$ | Law of Excluded Middle | (None) | |||
| 9 | 1, 2 | $q$ | Proof by Cases: $\text{PBC}$ | 8, 3 – 4, 5 – 7 | Assumptions 3 and 5 have been discharged |
$\blacksquare$