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Proof step 5 from proof rest-in (previous step)

n0 < # x, a ∈ restn(n0, x.tail), ∀ a, x. a ∈ restn(n0, x) ∧ n0 ≤ # x → a ∈ x ⊦ x = [], a ∈ x

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
x = a0 + y, n0 < # x, a ∈ restn(n0, y), ∀ a, x. a ∈ restn(n0, x) ∧ n0 ≤ # x → a ∈ x ⊦ x = [], a ∈ x

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x, a0, y]

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x, a0, y