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Proof step 15 from proof ordered-append (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x1, a1, y0]

Additional information

Used simplifier rules: le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);% ⊦ a + x = a ' + x; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x1, a1, y0