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Proof step 9 from proof sublist-move (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n + m ≤ m0, m0 ≤ # x0 + 1, n + m ≤ # x0 + 1, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ m0 = 0, firstn(m, restn(n, a ' + firstn(m0 - 1, x0))) = firstn(m, restn(n, a ' + x0))

Rule argument

Additional information

Used simplifier rules: c-07: ⊦ (a ' + x).rest = x; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list