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Proof step 7 from proof prefix-trans (previous step)

y ⊑ a ' + z0, a0 ' + y0 ⊑ y, y0 ⊑ y.tail ∧ y.tail ⊑ z0 → y0 ⊑ z0, ∀ x, y. x ⊑ y ∧ y ⊑ z0 → x ⊑ z0 ⊦ a0 ' + y0 ⊑ a ' + z0

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, y1]

Additional information

Used simplifier rules: prefix-nil: ⊦ [] ⊑ x;%prefix-empty: ⊦ x ⊑ [] ↔ x = [];%prefix-rec-03: ⊦ a ' + x ⊑ b ' + y ↔ a = b ∧ x ⊑ y;% ⊦ a + x = a ' + x; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y, a1, y1