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Proof step 10 from proof get-finto (previous step)

n0 ≤ # x0, m ≤ n0, m ≤ # x0, ∀ m, x0, a. m < n0 ∧ m ≤ # x0 ∧ n0 ≤ 1 + # x0 ∧ (∀ m, n. m < n ∧ m < # x0 ∧ n ≤ # x0 → (x0 to n)[m] = x0[m] ∨ x0 = []) → (a ' + x0 to n0)[m] = (a ' + x0)[m], ∀ m, n. m < n ∧ m < # x0 ∧ n ≤ # x0 → (x0 to n)[m] = x0[m] ∨ x0 = [] ⊦ (x0 to n0)[m - 1] = x0[m - 1], m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
m = m0 + 1, m ≤ # x0, m ≤ n0, n0 ≤ # x0, ∀ m, n. m < n ∧ m < # x0 ∧ n ≤ # x0 → (x0 to n)[m] = x0[m] ∨ x0 = [], ∀ m, x0, a. m < n0 ∧ m ≤ # x0 ∧ n0 ≤ 1 + # x0 ∧ (∀ m, n. m < n ∧ m < # x0 ∧ n ≤ # x0 → (x0 to n)[m] = x0[m] ∨ x0 = []) → (a ' + x0 to n0)[m] = (a ' + x0)[m] ⊦ (x0 to n0)[m0] = x0[m0], m = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m3]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m3