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Proof step 14 from proof sublist-prefix (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
x = a + y, n0 ≤ m, (m + 1) + m0 < # x, m + m0 < # x, firstn(n0, restn(m0, y)) ⊑ firstn(m, restn(m0, y)), ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], firstn(n0, restn(m0, y).rest) ⊑ firstn(m, restn(m0, y).rest)

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x, a, y] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a, y