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Proof step 17 from proof iterate-subseq (previous step)

((f ^ n0))(λ x. false) ⊆ f(((f ^ n0))(λ x. false)) → f(((f ^ n0))(λ x. false)) ⊆ f(f(((f ^ n0))(λ x. false))), ((f ^ n0))(λ x. false) ⊆ ((f ^ (n0 + 1)))(λ x. false) ⊦ f(((f ^ n0))(λ x. false)) ⊆ f(f(((f ^ n0))(λ x. false)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
((f ^ n0))(λ x. false) ⊆ ((f ^ (n0 + 1)))(λ x. false) ⊦ f(((f ^ n0))(λ x. false)) ⊆ f(f(((f ^ n0))(λ x. false))), ((f ^ n0))(λ x. false) ⊆ f(((f ^ n0))(λ x. false))

Proof step 17 from proof iterate-subseq (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information