Denying the Antecedent

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:

\(\ds p\)

\(\implies\)

\(\ds q\)

\(\ds \neg p\)

\(\)

\(\ds \)

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

\(\ds \neg q\)

\(\)

\(\ds \)

Corollary

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

Let its antecedent $p$ be false.

Then nothing can be inferred about the truth value of $q$.

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.

$\blacksquare$

Also see

• Affirming the Consequent
• Conditional and Inverse 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): denial of the antecedent
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): fallacy