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Proof step 63 from proof infinite-append-lem (previous step)

ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), ii(m) < ii(m0), m0 ≤ m, m < n0, m0 < n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦ m0 = n0 + 1, m = n0 + 1, n0 = m, n0 = 0, ii(m0) = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: m0

Additional information

Used simplifier rules: ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: m0