Modus Ponendo Ponens/Sequent Form/Proof by Truth Table
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Theorem
| \(\ds p\) | \(\implies\) | \(\ds q\) | ||||||||||||
| \(\ds p\) | \(\) | \(\ds \) | ||||||||||||
| \(\ds \vdash \ \ \) | \(\ds q\) | \(\) | \(\ds \) |
Proof
We apply the Method of Truth Tables.
$\begin{array}{|c|ccc||c|} \hline p & p & \implies & q & q\\ \hline \F & \F & \T & \F & \F \\ \F & \F & \T & \T & \T \\ \T & \T & \F & \F & \F \\ \T & \T & \T & \T & \T \\ \hline \end{array}$
As can be seen, when $p$ is true, and so is $p \implies q$, then $q$ is also true.
$\blacksquare$
Sources
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
- 1988: Alan G. Hamilton: Logic for Mathematicians (2nd ed.) ... (previous) ... (next): $\S 1$: Informal statement calculus: $\S 1.3$: Rules for manipulation and substitution: Proposition $1.9$