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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 (*).