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Proof step 39 from proof infinite-append-lem (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
ii(m) + n = n1 + ii(n0), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), 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) ⊦ lf(n0)[n1] = lf(m)[n], lf(m) = [], lf(n0) = [], m = n0 + 1, n0 = m

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ m + n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic