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Proof step 1 from proof cont-iterate-fix (conclusion)
cont(f) ⊦ (λ x0. ∃ n. ((f ^ n))(λ x. false)(x0)) = f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
cont(f) ⊦ (∃ n. ((f ^ n))(λ x. false)(x)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x)
Rule argument
Additional information