Universal Affirmative and Particular Negative are Contradictory

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Theorem

Consider the categorical statements:

\(\ds \mathbf A:\)    The universal affirmative:      \(\ds \forall x:\) \(\ds \map S x \implies \map P x \)      
\(\ds \mathbf O:\)    The particular negative:      \(\ds \exists x:\) \(\ds \map S x \land \neg \map P x \)      


Then $\mathbf A$ and $\mathbf O$ are contradictory.


Using the symbology of predicate logic:

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


Proof

\(\ds \) \(\) \(\ds \mathbf A\)
\(\ds \leadsto \ \ \) \(\ds \forall x: \, \) \(\ds \) \(\) \(\ds \map S x \implies \map P x\) Definition of $\mathbf A$
\(\ds \leadsto \ \ \) \(\ds \forall x: \, \) \(\ds \) \(\) \(\ds \neg \paren {\map S x \land \neg \map P x}\) Conditional is Equivalent to Negation of Conjunction with Negative
\(\ds \leadsto \ \ \) \(\ds \neg \exists x: \, \) \(\ds \) \(\) \(\ds \map S x \land \neg \map P x\) De Morgan's Laws: Denial of Existence
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds \neg \mathbf O\) Definition of $\mathbf O$


The argument reverses:

\(\ds \) \(\) \(\ds \mathbf O\)
\(\ds \leadsto \ \ \) \(\ds \exists x: \, \) \(\ds \) \(\) \(\ds \map S x \land \neg \map P x\) Definition of $\mathbf O$
\(\ds \leadsto \ \ \) \(\ds \exists x: \, \) \(\ds \) \(\) \(\ds \neg \paren {\map S x \implies \map P x}\) Conjunction with Negative is Equivalent to Negation of Conditional
\(\ds \leadsto \ \ \) \(\ds \neg \forall x: \, \) \(\ds \) \(\) \(\ds \map S x \implies \map P x\) De Morgan's Laws: Denial of Universality
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds \neg \mathbf A\) Definition of $\mathbf A$


The result follows by definition of contradictory.

$\blacksquare$


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