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:

\(\ds p\)

\(\implies\)

\(\ds q\)

\(\ds q\)

\(\)

\(\ds \)

\(\ds \not \vdash \ \ \)

\(\ds p\)

\(\)

\(\ds \)

This fallacy is called affirming the consequent.

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

• Denying the Antecedent
• Conditional and Converse are not Equivalent

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