# Affirming the Consequent

## Fallacy

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

Let its consequent $q$ be true.

Then it is a fallacy to assert that the antecedent $p$ is also necessarily true.

That is:

 $\ds p$ $\implies$ $\ds q$ $\ds q$  $\ds$ $\ds \not \vdash \ \$ $\ds p$  $\ds$

## Proof

We apply the Method of Truth Tables.

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

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

$\blacksquare$