Modus Tollendo Ponens/Sequent Form/Case 1

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Theorem

\(\ds p \lor q\) \(\) \(\ds \)
\(\ds \neg p\) \(\) \(\ds \)
\(\ds \vdash \ \ \) \(\ds q\) \(\) \(\ds \)


Proof 1

By the tableau method of natural deduction:

$p \lor q, \neg p \vdash q$
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

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