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Proof step 15 from proof eq (previous step)

ins≤(a, x0) = ins≤(a, y.tail), y.head < a, ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ y.head ' + x0 = y, y = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
ins≤(a, x0) = ins≤(a, y0), y = a0 + y0, a0 < a, ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, y = []

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [y, a0, y0]

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y, a0, y0