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Proof step 3 from proof ins-less-first (previous step)

ordered≤(x) ⊦ ins≤(a, x) = x.head ' + ins≤(a, x.tail), x = [], a < x.head

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
x = a0 + y, ordered≤(x) ⊦ ins≤(a, x) = a0 ' + ins≤(a, y), x = [], a < a0

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