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 $

$\ds \not \vdash \ \ $

$\ds p$

$\ds $

This fallacy is called .

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

1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation
1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$: Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): affirmation of the consequent
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): fallacy