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Proof step 16 from proof ordered-prefix (previous step)

ordered≤(y), x1 ⊑ y, (∃ x0. ordered≤(x0) ∧ x1 ⊑ x0) → ordered≤(x1) ⊦ x1 = [], y = [], y.head < a, ¬ x1.head < a ∧ ordered≤(x1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
y = a0 + y0, ordered≤(y), x1 ⊑ y, (∃ x0. ordered≤(x0) ∧ x1 ⊑ x0) → ordered≤(x1) ⊦ x1 = [], y = [], a0 < a, ¬ x1.head < a ∧ ordered≤(x1)

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