proposition: Lemma subseq_sat_rel : ∀ (a : ℕ → ℝ) (g : ℕ → ℕ → ℕ) (P : ℕ → ℝ → Prop), (∀ m N : ℕ, P m (a (g m N)) ) ⇒ ∀ k : ℕ, P k (a (create_seq g k)).
proof: Proof. Take a : (ℕ → ℝ). Take g : (ℕ → ℕ → ℕ). Take P : (ℕ → ℝ → Prop). Assume that (∀ m N : ℕ, P m (a (g m N))) (i). induction k. - We need to show that (P 0%nat (a (g 0%nat 0%nat))). By (i) we conclude that (P 0%nat (a (g 0%nat 0%nat))). - We need to show that (P (S k) (a (g (S k) (S (create_seq g k))))). By (i) we conclude that (P (S k) (a (g (S k) (S (create_seq g k))))). Qed.
imports: Coq.Reals.Reals ZArith Lra Classical Classical_Pred_Type ClassicalChoice Automation Libs.Negation Notations Tactics Waterprove