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Proof step 5 from proof firstN-dups-04 (previous step)

n0 < # x, ∀ x. dups(firstn(n0, x)) ∧ n0 ≤ # x → dups(x), x.head ∈ firstn(n0, x.tail) ∨ dups(firstn(n0, x.tail)) ⊦ x = [], dups(x)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

Additional information

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