Theorem 99: De Morgan’s laws for existentially quantified negations and negated universal quantification (inference rules)
Recall the usual notation for negation, existence statements and universal statements.
(a) Given that \exists x \neg \phi holds, we may infer that \neg \forall x \phi holds: \cfrac{\vdash \exists x \neg \phi}{\vdash \neg \forall x \phi}. (b) Given that \neg \forall x \phi holds, we may infer that \exists x \neg \phi holds: \cfrac{\vdash \neg \forall x \phi}{\vdash \exists x \neg \phi}.
Proof: In the below, let y be some variable which does not occur in \phi , and note that by [27], it follows that x is freely substitutable for y in \phi .
(a): Assume \forall x \phi . Using universal elimination, we infer that the variable substitution [\phi | x \to y] holds. Then, by double negation introduction, we infer that the double negation \neg \neg [\phi | x \to y] holds; by negation elimination, we therefore get that \neg [\phi | x \to y] entails \neg \exists x \neg \phi .
Using existential elimination (with the premise \exists x \neg \phi and the entailment we just found), we infer that \exists x \phi holds. Therefore, \forall x \phi entails \exists x \phi . But by weakening, since \vdash \exists x \neg \phi is a premise, we infer that \forall x \phi entails \exists x \neg \phi . We conclude, using negation introduction, that \neg \forall x \phi holds.
(b): Assume \neg \exists x \neg \phi . Using part (b) of [98], we infer \forall x \neg \neg \phi holds. Thus, by universal elimination, [\neg \neg \phi | x \to y] holds. This is the same as \neg \neg [\phi | x \to y ] , so we get by double negation elimination that [\phi | x \to y] holds. By universal introduction, it follows that \forall x \phi holds. Thus, \neg \exists x \neg \phi entails \forall x \phi .
But by weakening, \neg \exists x \neg \phi also entails \neg \forall x \phi ; we therefore infer using negation introduction that \neg \neg \neg \exists x \neg \phi holds. By double negation elimination, we conclude that \exists x \neg \phi holds. \square