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Proof step 6 from proof get-finto (previous step)

m < n0 +1, n0 +1 ≤ 1 + # x0, m ≤ # 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 +1)[m] = (a ' + x0)[m]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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 = [] ⊦ (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% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ m +1 = m + 1; (s)