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Proof step 13 from proof eq (previous step)

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

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] → [y, a1, y0] flags:

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%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%irreflexivity: ⊦ ¬ a < a; from specification oelem%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: y, a1, y0