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Proof step 9 from proof iterate-is-fix (previous step)

cont(f), f(λ x. ∃ n. ((f ^ n))(λ x. false)(x))(x0) ⊦ ∃ n. ((f ^ n))(λ x. false)(x0), ∃ n. ((f ^ n))(λ x. false)(x)

User applied rule insert lemma

And obtained the following new goals

Proof step 10:
cont(f) ⊦ (λ x0. ∃ n. ((f ^ n))(λ x. false)(x0)) = f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))
Proof step 11:
(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0)) = f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0)), cont(f), f(λ x. ∃ n. ((f ^ n))(λ x. false)(x))(x0) ⊦ ∃ n. ((f ^ n))(λ x. false)(x0), ∃ n. ((f ^ n))(λ x. false)(x)

Rule argument

spec: inst: name: cont-iterate-fix seq: cont(f) ⊦ (λ x0. ∃ n. ((f ^ n))(λ x. false)(x0)) = f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0)) subst: [f] → [f] flags: current, []

Additional information

Used terms: f