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Proof step 5 from proof get-finto (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 ≤ # x0, m ≤ n0, m ≤ # x0, ∀ m, x0, a. m ≤ # x0 ∧ n0 ≤ # x0 + 1 ∧ m < n0 ∧ (∀ m, n. m < n ∧ n ≤ # x0 ∧ m < # x0 → (x0 to n)[m] = x0[m] ∨ n = 0 ∨ x0 = []) → (a ' + x0 to n0)[m] = (a ' + x0)[m] ∨ n0 = 0, ∀ m, n. m < n ∧ n ≤ # x0 ∧ m < # x0 → (x0 to n)[m] = x0[m] ∨ n = 0 ∨ x0 = [] ⊦ (a ' + (x0 to n0))[m] = (a ' + x0)[m]

Rule argument

Additional information

Used simplifier rules: to-succ-cons-l-lem: ⊦ a ' + x to n + 1 = a + (x to n);%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2