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Proof step 11 from proof del1 (previous step)

ordered<(a0 ' + x0), ∀ a1. ordered<(x0 -1l a1) ⊦ x0 -1l a = [], a = a0, a0 < (x0 -1l a).head ∧ ordered<(x0 -1l a)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
x0 -1l a = a2 + y, ordered<(a0 ' + x0), ∀ a1. ordered<(x0 -1l a1) ⊦ x0 -1l a = [], a = a0, a0 < a2 ∧ ordered<(x0 -1l a)

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x0 -1l a, a2, y]

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x0 -1l a, a2, y