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Proof step 20 from proof sublist-move (previous step)

0 < n, m1 ≤ # x0, n + m ≤ m1 + 1, n + m ≤ 1 + # x0, ∀ n0, m0, n, m. n0 + m0 ≤ # x0 ∧ n + m ≤ m0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, restn(n -1, (a ' + firstn(m1, x0)).tail)) = firstn(m, restn(n, a ' + x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
n + m ≤ 1 + # x0, n + m ≤ m1 + 1, m1 ≤ # x0, ∀ n0, m0, n, m. n0 + m0 ≤ # x0 ∧ n + m ≤ m0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, restn(n - 1, firstn(m1, x0))) = firstn(m, restn(n, a ' + x0)), n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < m ↔ m ≠ 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ n -1 = n - 1; (s)