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Proof step 8 from proof ole-01 (previous step)

a0 < a, ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ ins≤(a, x0).first < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ordered≤(a0 ' + x0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
ins≤(a, x0) = a1 + y ∧ ins≤(a, x0) ≠ [], a0 < a, ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ordered≤(a0 ' + x0)

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] → [ins≤(a, x0), a1, y] flags:

Additional information

Used simplifier rules: empty: ⊦ ins≤(a, x) ≠ [];%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: ins≤(a, x0), a1, y