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Proof step 4 from proof firstrestex (previous step)

n0 + 1 < # x, ∀ x. n0 < # x → x = firstn(n0, x) + restn(n0, x) ∨ x = [] ⊦ x = [], x = (x.first ' + firstn(n0, x.rest)) + restn(n0, x.rest)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
x = a + y, n0 + 1 < # x, ∀ x. n0 < # x → x = firstn(n0, x) + restn(n0, x) ∨ x = [] ⊦ x = [], x = (a ' + firstn(n0, y)) + restn(n0, y)

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