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Proof step 42 from proof infinite-append-lem (previous step)

ii(m) + n = n1 + ii(n0), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii((n0 - 1) + 1) = ii(n0 - 1) + # lf(n0 - 1), (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(n0 - 1)) to # lf(n0 - 1) = lf(n0 - 1), m < n0, n < # lf(m), n1 < # lf(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. m0 < n0 → ii(m0) ≤ ii(n0 - 1) ⊦ lf(n0)[n1] = 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 43:
ii(m) + n = n1 + ii(n0), ii(m1 + 1) = ii(m1) + # lf(m1), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, n0 = m1 + 1, (nf from ii(m1)) to # lf(m1) = lf(m1), (nf from ii(m)) to # lf(m) = lf(m), n1 < # lf(n0), n < # lf(m), m < n0, ∀ m0. m0 < n0 → ii(m0) ≤ ii(m1), ∀ 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, n0 = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n0, m1]

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ m < 0; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m1