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Proof step 18 from proof sublist-append-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
n + n0 + n1 ≤ # x0 + 1, ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ firstn(n0, restn(n - 1, x0)) + firstn(n1, restn(n + n0, a ' + x0)) = firstn(n0 + n1, restn(n - 1, x0)), n = 0

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat