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Proof step 9 from proof get-finto (previous step)

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] ⊦ (a ' + (x0 to n0))[m] = x0[m - # a '], m < # a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
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

Rule argument

Additional information

Used simplifier rules: ⊦ m < 1 ↔ m = 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s)