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Proof step 5 from proof ordered-append (previous step)

ordered≤(a ' + x1), ordered≤(x0), x0 ≠ [] → ¬ x0.head < (a ' + x1).last, ∀ x. ordered≤(x) ∧ (x1 ≠ [] ∧ x ≠ [] → ¬ x.head < x1.last) → ordered≤(x1 + x) ⊦ ordered≤((a ' + x1) + x0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
x0 = [], ordered≤(x0), ordered≤(a ' + x1), x0 ≠ [] → ¬ x0.head < (a ' + x1).last, ∀ x. ordered≤(x) ∧ (x1 ≠ [] ∧ x ≠ [] → ¬ x.head < x1.last) → ordered≤(x1 + x) ⊦ ordered≤((a ' + x1) + x0)
Proof step 7:
x0 = a0 + y, ordered≤(x0), ordered≤(a ' + x1), x0 ≠ [] → ¬ a0 < (a ' + x1).last, ∀ x. ordered≤(x) ∧ (x1 ≠ [] ∧ x ≠ [] → ¬ x.head < x1.last) → ordered≤(x1 + x) ⊦ x0 = [], ordered≤((a ' + x1) + x0)

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, a0, y]

Additional information

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