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Proof step 22 from proof eq (previous step)

a ' + a0 ' + x0 = ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, y = [], a0 < a, y.head < a

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] → [y, a1, y0]

Additional information

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