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Proof step 5 from proof iterate-is-fix (previous step)

f(λ x. ∃ n. ((f ^ n))(λ x. false)(x)) ⊆ (λ x. ∃ n. ((f ^ n))(λ x. false)(x)) → (∃ n. ((f ^ n))(λ x. false)(x)), cont(f) ⊦ ∃ n. ((f ^ n))(λ x. false)(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
cont(f) ⊦ f(λ x. ∃ n. ((f ^ n))(λ x. false)(x)) ⊆ (λ x. ∃ n. ((f ^ n))(λ x. false)(x)), ∃ n. ((f ^ n))(λ x. false)(x)

Proof step 5 from proof iterate-is-fix (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information