pub proof fn lemma_full_good_set_implies_forall<T>(
p: FnSpec<(T,), bool>,
q: FnSpec<(T,), bool>,
)Expand description
requires
Set::new(|x: T| p(x)).finite(),Set::new(|x: T| p(x)).len() == Set::new(|x: T| p(x)).filter(q).len(),ensuresforall |x: T| #[trigger] p(x) ==> q(x),If all elements in the finite set Set::new(|x: T| p(x)) satisfy the predicate q, then all elements
satisfying p also satisfy q.