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Proof step 16 from proof iterate-subseq (previous step)

((f ^ n0))(λ x. false) ⊆ ((f ^ (n0 + 1)))(λ x. false), ∀ S, S0. S ⊆ S0 → f(S) ⊆ f(S0) ⊦ f(((f ^ n0))(λ x. false)) ⊆ f(f(((f ^ n0))(λ x. false)))

User applied rule all left

And obtained the following new goals

Proof step 17:
((f ^ n0))(λ x. false) ⊆ ((f ^ (n0 + 1)))(λ x. false), ((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(((f ^ n0))(λ x. false)) ⊆ f(f(((f ^ n0))(λ x. false)))

Rule argument

pos: left 2 terms: ((f ^ n0))(λ x. false), f(((f ^ n0))(λ x. false)) abort: F discard: T computed: T

Additional information

Used terms: ((f ^ n0))(λ x. false), f(((f ^ n0))(λ x. false))