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Proof step 7 from proof app (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
x0 = a0 + y0, ordered≤((a ' + x0) + y), (∃ y. ordered≤(x0 + y)) → ordered≤(x0) ⊦ x0 = [], ¬ a0 < a ∧ ordered≤(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, a0, y0] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x0, a0, y0