pub proof fn lemma_empty_bad_set_implies_forall<T>(
p: FnSpec<(T,), bool>,
q: FnSpec<(T,), bool>,
)Expand description
requires
Set::new(|x: T| p(x)).filter(|x| !q(x)).is_empty(),ensuresforall |x: T| #[trigger] p(x) ==> q(x),If no element in set Set::new(|x: T| p(x)) satisfies the predicate q, then all elements
satisfying p also satisfy q.