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Proof step 5 from proof get-from-01 (previous step)

m1 +1 + m0 ≤ # x0, ∀ m, m0. m + m0 < # x0 → (x0 from m)[m0] = x0[m + m0] ∨ x0 = [], ∀ m0, x0, a. m1 + m0 ≤ # x0 ∧ (∀ m, m0. m + m0 < # x0 → (x0 from m)[m0] = x0[m + m0] ∨ x0 = []) → (a ' + x0 from m1)[m0] = (a ' + x0)[m1 + m0] ⊦ (a ' + x0 from m1 +1)[m0] = (a ' + x0)[m1 +1 + m0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m1 + m0 < # x0, ∀ m0, x0, a. m1 + m0 ≤ # x0 ∧ (∀ m, m0. m + m0 < # x0 → (x0 from m)[m0] = x0[m + m0] ∨ x0 = []) → (a ' + x0 from m1)[m0] = (a ' + x0)[m1 + m0], ∀ m, m2. m + m2 < # x0 → (x0 from m)[m2] = x0[m + m2] ⊦ (x0 from m1)[m0] = x0[m1 + m0], x0 = []

Rule argument

Additional information

Used simplifier rules: from-succ-cons-l-lem: ⊦ a ' + x from n + 1 = x from n;% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ (a ' + x)[n + 1] = x[n]; (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% ⊦ m +1 = m + 1; (s)