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Proof step 1 from proof infinite-append-lem (conclusion)

⊦ ∀ n. ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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