Modus Tollendo Ponens/Sequent Form/Case 2
Jump to navigation
Jump to search
Theorem
\(\ds p \lor q\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \neg q\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds p\) | \(\) | \(\ds \) |
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \lor q$ | Premise | (None) | ||
2 | 2 | $\neg q$ | Premise | (None) | ||
3 | 3 | $q$ | Assumption | (None) | ||
4 | 2 | $q \implies p$ | Sequent Introduction | 2 | False Statement implies Every Statement | |
5 | 2, 3 | $p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 4, 3 | ||
6 | 6 | $p$ | Assumption | (None) | ||
7 | 1, 2 | $p$ | Proof by Cases: $\text{PBC}$ | 1, 3 – 5, 6 – 6 | Assumptions 3 and 6 have been discharged |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Theorem $53$
- 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 2 \ \text{(c)}$