Conversion per Accidens

Theorem

Consider the categorical statements:

		\(\ds \map {\mathbf A} {S, P}:\)	The universal affirmative:	\(\ds \forall x:\)	\(\ds \map S x \implies \map P x \)
		\(\ds \map {\mathbf I} {P, S}:\)	The particular affirmative:	\(\ds \exists x:\)	\(\ds \map P x \land \map S x \)

Then:

$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {P, S}$

if and only if:

$\exists x: \map S x$

Using the symbology of predicate logic:

$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map P x \land \map S x} }$

This law has the traditional name conversion per accidens of $\mathbf A$.

Thus the $\mathbf A$ form converts per accidens to the $\mathbf I$ form.

Proof

From Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous:

$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$

From Law of Simple Conversion of I:

$\paren {\exists x: \map S x \land \map P x} \implies \paren {\exists x: \map P x \land \map S x}$

Hence the result.

$\blacksquare$

Also defined as

Some sources gloss over the possibility of $\map S x$ being vacuous and merely report that the universal affirmative $\map {\mathbf A} {S, P}$ implies the particular affirmative $\map {\mathbf I} {P, S}$.

Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism