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

m = m0 + 1, 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 = [] ⊦ (x0 to n0)[m0] = x0[m0], m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2