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Proof step 13 from proof ole-01 (previous step)

ins≤(a, x0) = a1 ' + y, a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ¬ x0.first < a0 ∧ ordered≤(x0), x0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
x0 = a2 + y0, ins≤(a, x0) = a1 ' + y, a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ¬ a2 < a0 ∧ ordered≤(x0), 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] → [x0, a2, y0] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x0, a2, y0