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Proof step 4 from proof get-finto (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ 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

Rule argument

indvar: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;