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Proof step 17 from proof ordered-prefix (previous step)

ordered≤(a0 ' + y0), x1 ⊑ a0 ' + y0, (∃ x0. x1 ⊑ x0 ∧ ordered≤(x0)) → ordered≤(x1) ⊦ x1 = [], a0 < a, ¬ x1.first < a ∧ ordered≤(x1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
x1 = a1 + y, ordered≤(a0 ' + y0), x1 ⊑ a0 ' + y0, (∃ x0. x1 ⊑ x0 ∧ ordered≤(x0)) → ordered≤(x1) ⊦ x1 = [], a0 < a, ¬ a1 < a ∧ ordered≤(x1)

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x1, a1, y] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x1, a1, y