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Proof step 13 from proof ordered-append (previous step)

ordered≤(a ' + x1), ordered≤(x1 + a0 ' + y), x1.first < a, ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.first < 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 inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x1, a1, y0] flags:

Additional information

Used simplifier rules: le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);%c: ⊦ a + x = a ' + x; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x1, a1, y0