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Proof step 10 from proof sublist-move (previous step)

n + m ≤ 1 + # x0, m0 ≤ 1 + # x0, n + m ≤ m0, ∀ 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, (a ' + x0).head ' + firstn(m0 -1, (a ' + x0).tail))) = firstn(m, restn(n, a ' + x0)), m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n + m ≤ m0, m0 ≤ 1 + # x0, 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, a ' + firstn(m0 - 1, x0))) = firstn(m, restn(n, a ' + x0)), m0 = 0

Rule argument

Additional information

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