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Proof step 13 from proof eq (previous step)

a0 ' + ins≤(a, x0) = ins≤(a, y), a0 < a, a0 ∈ ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, y = [], 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)% ⊦ a ' + x ≠ []; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem% ⊦ ¬ a < a; (s)%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y, a1, y0