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Proof step 5 from proof iterate-subseq-leq (previous step)

∀ f, n. monotone(f) ∧ m0 ≤ n → ((f ^ m0))(λ x. false) ⊆ ((f ^ n))(λ x. false), monotone(f), m0 +1 ≤ n ⊦ ((f ^ m0 +1))(λ x. false) ⊆ ((f ^ n))(λ x. false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
monotone(f), m0 < n, ∀ f, n. monotone(f) ∧ m0 ≤ n → ((f ^ m0))(λ x. false) ⊆ ((f ^ n))(λ x. false) ⊦ n = 0, ((f ^ (m0 + 1)))(λ x. false) ⊆ ((f ^ n))(λ x. false)

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2