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Proof step 3 from proof get-finto (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m ≤ # x0, n ≤ # x0 + 1, m < n, ∀ m, n. m < n ∧ n ≤ # x0 ∧ m < # x0 → (x0 to n)[m] = x0[m] ∨ n = 0 ∨ x0 = [] ⊦ (a ' + x0 to n)[m] = (a ' + x0)[m], n = 0

Rule argument

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list