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Proof step 8 from proof sublist-prefix (previous step)

n0 < n1, n1 + m0 < # x, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail) ⊦ x = [], firstn(n0, restn(m0, x).tail.tail) ⊑ firstn(n1 - 1, restn(m0, x).tail.tail)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
n1 = m + 1, n1 + m0 < # x, n0 < n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail) ⊦ n1 = 0, x = [], firstn(n0, restn(m0, x).tail.tail) ⊑ firstn(m, restn(m0, x).tail.tail)

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n1, m]

Additional information

Used simplifier rules: ⊦ ¬ m < 0; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, m