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Proof step 17 from proof del1-01-02 (previous step)

a < a0, ordered≤(x0), ∀ a1, b0. ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ∨ a1 = b0 ⊦ a = a0, ins≤(a, x0) = a ' + x0, x0 = [], x0.first < a0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
x0 = a2 + y, a < a0, ordered≤(x0), ∀ a1, b0. ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ∨ a1 = b0 ⊦ a = a0, ins≤(a, x0) = a ' + x0, x0 = [], a2 < a0

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, y] flags:

Additional information

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