proposition: Lemma index_seq_equiv (n : ℕ → ℕ) : is_index_seq n ⇔ is_index_sequence n.
proof: Proof. We show both directions. - We need to show that (is_index_seq n ⇨ is_index_sequence n). intro. unfold is_index_sequence. Take k : ℕ. unfold is_index_seq in H. We conclude that ((n k < n (k + 1))%nat). - We need to show that (is_index_sequence n ⇨ is_index_seq n). intro. unfold is_index_seq. Take k : ℕ. unfold is_index_sequence in H. We conclude that ((n k < n (k + 1))%nat). Qed.
imports: Coq.Reals.Reals Automation Libs.Analysis.MetricSpaces Libs.Analysis.SequencesMetric Notations Tactics