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Proof step 14 from proof sublist-append-04 (previous step)

n0 ≠ 0, n0 + n1 = m + 1, m ≤ # x0, ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ ((a ' + x0).first ' + firstn(n0 -1, (a ' + x0).rest)) + firstn(n1, restn(n0, a ' + x0)) = a ' + firstn(m, x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
n0 + n1 = m + 1, m ≤ # x0, ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ n0 = 0, firstn(n0 - 1, x0) + firstn(n1, restn(n0, a ' + x0)) = firstn(m, x0)

Rule argument

Additional information

Used simplifier rules: app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%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