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

∀ S, S0. S ⊆ S0 → f(S) ⊆ f(S0), m0 ≤ n0, ((f ^ m0))(λ x. false) ⊆ ((f ^ n0))(λ x. false) ⊦ f(((f ^ m0))(λ x. false)) ⊆ f(((f ^ n0))(λ x. false))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: ((f ^ m0))(λ x. false), ((f ^ n0))(λ x. false)

Additional information

Used terms: ((f ^ m0))(λ x. false), ((f ^ n0))(λ x. false)