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Proof step 38 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), m < n0, n < # lf(m), ii(m) + n < ii(n0) + # lf(n0), ii(n0) ≤ ii(m) + n, ∀ 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) ⊦ lf(n0)[ii(m) + n - ii(n0)] = lf(m)[n], lf(m) = [], lf(n0) = [], m = n0 + 1, n0 = m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 39:
ii(m) + n = n1 + ii(n0), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), ii(m) + n < ii(n0) + # lf(n0), n < # lf(m), m < n0, ii(n0) ≤ ii(m) + n, ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ 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)) ⊦ lf(n0)[n1] = lf(m)[n], lf(n0) = [], lf(m) = [], m = n0 + 1, n0 = m

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [ii(m) + n, ii(n0), n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: ii(m) + n, ii(n0), n1