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Proof step 2 from proof ols-rec (previous step)

⊦ ordered<(x) ↔ x.first < x.rest.first ∧ ordered<(x.rest), x.rest = [], x = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
x = a + y ⊦ ordered<(x) ↔ a < y.first ∧ ordered<(y), y = [], x = []

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] → [x, a, y] flags:

Additional information

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