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Proof step 23 from proof max-ins-max (previous step)

max(ins≤(a, x0)) < a0, a0 < a, max(ins≤(a, x0)) < a, ∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ a0 = a, x0 = [], a < (max(x0) < a0 ⊃ a0;max(x0)), ¬ x0.first < a ∧ ordered≤(x0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
x0 = a1 + y, max(ins≤(a, x0)) < a0, a0 < a, max(ins≤(a, x0)) < a, ∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ a0 = a, x0 = [], a < (max(x0) < a0 ⊃ a0;max(x0)), ¬ a1 < 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, a1, y] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x0, a1, y