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Proof step 19 from proof sublist-append-04 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
n = m + 1, 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(m, x0)) + firstn(n1, restn(n + n0, a ' + x0)) = firstn(n0 + n1, restn(m, x0)), n = 0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m