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Proof step 21 from proof cont-iterate-fix (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
monotone(f), ((f ^ (n0 + 1)))(λ x. false)(x), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ ∃ n. f(((f ^ n))(λ x. false))(x), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x)

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2