Conditional and Converse are not Equivalent
Jump to navigation
Jump to search
This article is a landmark page. It was the 9000th proof on $\mathsf{Pr} \infty \mathsf{fWiki}$!
This article is a landmark page. It was the 9000th proof on $\mathsf{Pr} \infty \mathsf{fWiki}$!Theorem
- $p \implies q$
is not logically equivalent to its converse:
- $q \implies p$
Proof
We apply the Method of Truth Tables to the proposition:
- $\paren {p \implies q} \iff \paren {q \implies p}$
$\begin{array}{|ccc|c|ccc|} \hline p & \implies & q) & \iff & (q & \implies & p) \\ \hline \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \T & \F & \T & \F & \F \\ \T & \F & \F & \F & \F & \T & \T \\ \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations.
$\blacksquare$
Also see
Sources
- 1946: Alfred Tarski: Introduction to Logic and to the Methodology of Deductive Sciences (2nd ed.) ... (previous) ... (next): $\S \text{II}.14$: Application of laws of sentential calculus in inference
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.5$: Theorems and Proofs
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(3)$