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Proof step 6 from proof iterate-subseq-leq (previous step)

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)

User applied rule constructor cut

And obtained the following new goals

Proof step 7:
n = n0 +1, 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

var: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;