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Proof step 18 from proof del1 (previous step)

x0 -1l a = a2 ' + y, a0 < x0.head, a2 < x0.head, ordered<(x0 -1l a), ordered<(x0), a2 ∈ x0 -1l a, a2 ∈ x0 ⊦ x0 -1l a = [], a = a0, x0 = [], a0 < a2

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] → [x0, a1, y0]

Additional information

Used simplifier rules: orderedls-ls: ⊦ b < a ∧ b ∈ x → ¬ ordered<(a ' + x);% ⊦ a + x = a ' + x; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x0, a1, y0