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We show for arbitrary formula A

(~forall x: A) => (exists x: ~A)
by showing (contraposition)
(~exists x: ~A) => (~~forall x: A)
i.e. (propositional consequence and substitution, see next subsection)
(~exists x: ~A) => (forall x: A).
We assume (*) ~exists x: ~A and show forall x: A. Take arbitrary and assume ~A. Then we have (exists x: ~A) which contradicts (*).
Author: Wolfgang Schreiner
Last Modification: November 30, 1999

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