Law of Simple Conversion of E
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Theorem
Consider the universal negative categorical statement No $S$ is $P$:
- $\map {\mathbf E} {S, P}: \forall x: \map S x \implies \neg \map P x$
Then No $P$ is $S$:
- $\map {\mathbf E} {P, S}$
Proof
\(\ds \) | \(\) | \(\ds \map {\mathbf E} {S, P}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds \map S x \implies \neg \map P x\) | Definition of Universal Negative | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds \neg \paren {\map S x \land \map P x}\) | Modus Ponendo Tollens | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds \neg \paren {\map P x \land \map S x}\) | Conjunction is Commutative | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall x: \, \) | \(\ds \) | \(\) | \(\ds \map P x \implies \neg \map S x\) | Modus Ponendo Tollens | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds \map {\mathbf E} {P, S}\) | Definition of Universal Negative |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism