Denying the Antecedent/Corollary

Corollary to Denying the Antecedent

Let $p \implies q$ be a conditional statement.

Let its antecedent $p$ be false.

Then nothing can be inferred about the truth value of $q$.


Proof

By applying the Method of Truth Tables to the proposition, it is seen that $q$ can be either true or false, and both are consistent with both $\neg p$ and $p \implies q$.

$\begin{array}{|ccc|cc||cc|} \hline p & \implies & q & \neg & p & \neg & q\\ \hline \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \T & \T & \F & \F & \T \\ \T & \F & \F & \F & \T & \T & \F \\ \T & \T & \T & \F & \T & \F & \T \\ \hline \end{array}$

$\blacksquare$


Sources

  • 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): Chapter $1$: Sets, Functions, and Relations: $\S 2$: Some Remarks on the Use of the Connectives and, or, implies
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