# Denying the Antecedent

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## Fallacy

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

Let its antecedent $p$ be false.

Then it is a fallacy to assert that the consequent $q$ is also necessarily false.

That is:

$p \implies q, \neg p \not \vdash \neg q$

## Proof

We apply the Method of Truth Tables to the proposition.

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

As can be seen, when $p \implies q$ is true, and so is $\neg p$, then it is not always the case that $\neg q$ is also true.

$\blacksquare$