Modus Tollendo Ponens/Sequent Form/Case 1
Theorem
| \(\ds p \lor q\) | \(\) | \(\ds \) | ||||||||||||
| \(\ds \neg p\) | \(\) | \(\ds \) | ||||||||||||
| \(\ds \vdash \ \ \) | \(\ds q\) | \(\) | \(\ds \) |
Proof 1
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \lor q$ | Premise | (None) | ||
| 2 | 2 | $\neg p$ | Premise | (None) | ||
| 3 | 3 | $p$ | Assumption | (None) | ||
| 4 | 2 | $p \implies q$ | Sequent Introduction | 2 | False Statement implies Every Statement | |
| 5 | 2, 3 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 4, 3 | ||
| 6 | 6 | $q$ | Assumption | (None) | ||
| 7 | 1, 2 | $q$ | Proof by Cases: $\text{PBC}$ | 1, 3 – 5, 6 – 6 | Assumptions 3 and 6 have been discharged |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables to the proposition.
$\begin{array}{|ccc|cc||c|} \hline p & \lor & q & \neg & p & q\\ \hline \F & \F & \F & \T & \F & \F \\ \F & \T & \T & \T & \F & \T \\ \T & \T & \F & \F & \T & \F \\ \T & \T & \T & \F & \T & \T \\ \hline \end{array}$
As can be seen, when $p \lor q$ is true, and so is $\neg p$, then $q$ is also true.
$\blacksquare$
Sources
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $4.$
- 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $5$
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): $\S 1.14$: Exercise $17 \ \text{(iii)}$
- 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 $1.5: \ 1 \ \text{(c)}$