Affirming the Consequent

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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:

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


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.

$\blacksquare$


Also see


Sources