Universal Affirmative and Negative are both False iff Particular Affirmative and Negative are both True
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Theorem
Consider the categorical statements:
| The universal affirmative: | \(\ds \map {\mathbf A} {S, P}:\) | \(\ds \forall x: \map S x \) | \(\ds \implies \) | \(\ds \map P x \) | |||||
| The universal negative: | \(\ds \map {\mathbf E} {S, P}:\) | \(\ds \forall x: \map S x \) | \(\ds \implies \) | \(\ds \neg \map P x \) | |||||
| The particular affirmative: | \(\ds \map {\mathbf I} {S, P}:\) | \(\ds \exists x: \map S x \) | \(\ds \land \) | \(\ds \map P x \) | |||||
| The particular negative: | \(\ds \map {\mathbf O} {S, P}:\) | \(\ds \exists x: \map S x \) | \(\ds \land \) | \(\ds \neg \map P x \) |
Then:
- $\map {\mathbf A} {S, P}$ and $\map {\mathbf E} {S, P}$ are both false
- $\map {\mathbf I} {S, P}$ and $\map {\mathbf O} {S, P}$ are both true.
Proof
Necessary Condition
Let $\map {\mathbf A} {S, P}$ and $\map {\mathbf E} {S, P}$ both be false.
| \(\text {(1)}: \quad\) | \(\ds \neg \map {\mathbf A} {S, P}\) | \(\land\) | \(\ds \neg \map {\mathbf E} {S, P}\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \neg \map {\mathbf A} {S, P}\) | \(\) | \(\ds \) | Rule of Simplification: from $(1)$ | |||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map {\mathbf O} {S, P}\) | \(\) | \(\ds \) | Universal Affirmative and Particular Negative are Contradictory: from $(2)$ | |||||||||
| \(\text {(4)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \neg \map {\mathbf E} {S, P}\) | \(\) | \(\ds \) | Rule of Simplification: from $(1)$ | |||||||||
| \(\text {(5)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map {\mathbf I} {S, P}\) | \(\) | \(\ds \) | Particular Affirmative and Universal Negative are Contradictory: from $(4)$ | |||||||||
| \(\text {(6)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map {\mathbf I} {S, P}\) | \(\land\) | \(\ds \map {\mathbf O} {S, P}\) | Rule of Conjunction: from $(5)$ and $(3)$ |
Hence $\map {\mathbf I} {S, P}$ and $\map {\mathbf O} {S, P}$ are both true.
$\Box$
Sufficient Condition
Let $\map {\mathbf I} {S, P}$ and $\map {\mathbf O} {S, P}$ both be true.
| \(\text {(1)}: \quad\) | \(\ds \map {\mathbf I} {S, P}\) | \(\land\) | \(\ds \map {\mathbf O} {S, P}\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map {\mathbf I} {S, P}\) | \(\) | \(\ds \) | Rule of Simplification: from $(1)$ | |||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \neg \map {\mathbf E} {S, P}\) | \(\) | \(\ds \) | Particular Affirmative and Universal Negative are Contradictory: from $(2)$ | |||||||||
| \(\text {(4)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map {\mathbf O} {S, P}\) | \(\) | \(\ds \) | Rule of Simplification: from $(1)$ | |||||||||
| \(\text {(5)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \neg \map {\mathbf A} {S, P}\) | \(\) | \(\ds \) | Universal Affirmative and Particular Negative are Contradictory: from $(4)$ | |||||||||
| \(\text {(6)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \neg \map {\mathbf A} {S, P}\) | \(\land\) | \(\ds \neg \map {\mathbf E} {S, P}\) | Rule of Conjunction: from $(5)$ and $(3)$ |
Hence $\map {\mathbf A} {S, P}$ and $\map {\mathbf E} {S, P}$ are both false.
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism