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Proof step 17 from proof sublist-append-04 (previous step)

n0 + n1 = m + 1, n0 = m0 + 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)) ⊦ firstn(m0, x0) + firstn(n1, restn(n0, a ' + x0)) = firstn(m, x0), n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
m0 + n1 ≤ # 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)) ⊦ firstn(m0, x0) + firstn(n1, restn(m0, x0)) = firstn(m0 + n1, x0)

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.tail);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ m + 1 ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)