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Proof step 8 from proof ordered-max (previous step)

max(x0) = x0.last, ordered≤(x0), x0.last < a ⊦ x0 = [], a = x0.last, x0.first < a

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
x0 = a0 + y, max(x0) = x0.last, ordered≤(x0), x0.last < a ⊦ x0 = [], a = x0.last, a0 < a

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, y] flags:

Additional information

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