Rule of Transposition/Formulation 2/Proof by Truth Table
Jump to navigation
Jump to search
Theorem
- $\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$
Proof
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc|c|ccccc|} \hline p & \implies & q) & \iff & (\neg & q & \implies & \neg & p) \\ \hline \F & \T & \F & \T & \T & \F & \T & \T & \F \\ \F & \T & \T & \T & \F & \T & \T & \T & \F \\ \T & \F & \F & \T & \T & \F & \F & \F & \T \\ \T & \T & \T & \T & \F & \T & \T & \F & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{I}: 13$: Logical Equivalences: Exercise $\text{(A)} \ 7$