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Proof step 2 from proof k-dups (previous step)

dups(x) ⊦ x = [], (x.head ∈ x.tail ∨ dups(x.tail)) ∧ # x ≠ 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
x = a + y, dups(x) ⊦ x = [], (a ∈ y ∨ dups(y)) ∧ # x ≠ 1

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