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Proof step 20 from proof cont-iterate-fix (previous step)

((f ^ n))(λ x. false)(x), monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x), ∃ n. f(((f ^ n))(λ x. false))(x), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x)

User applied rule constructor cut

And obtained the following new goals

Proof step 21:
n = n0 +1, ((f ^ n))(λ x. false)(x), monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x), ∃ n. f(((f ^ n))(λ x. false))(x), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x)

Rule argument

var: n

Additional information

Used simplifier rules: iterate-base: ⊦ ((f ^ 0))(S) = S;%nat freely generated by 0, +1;;