Modus Ponendo Ponens/Variant 3/Proof by Truth Table
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Theorem
- $\vdash \paren {\paren {p \implies q} \land p} \implies q$
Proof
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
$\begin{array}{|ccccc|c|c|} \hline ((p & \implies & q) & \land & p) & \implies & q \\ \hline \F & \T & \F & \F & \F & \T & \F \\ \F & \T & \T & \F & \F & \T & \T \\ \T & \F & \F & \F & \T & \T & \F \\ \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{I}: 10$: Testing the Validity of Arguments
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic