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Proof step 6 from proof prefix-trans (previous step)

y ⊑ a ' + z0, a0 ' + y0 ⊑ y, y0 ⊑ y.rest ∧ y.rest ⊑ 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 inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [y, a1, y1] flags:

Additional information

Used simplifier rules: prefix-nil: ⊦ [] ⊑ x;%prefix-empty: ⊦ x ⊑ [] ↔ x = [];%prefix-rec-03: ⊦ a ' + x ⊑ b ' + y ↔ a = b ∧ x ⊑ y;%c: ⊦ a + x = a ' + x; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list%e: ⊦ a ' ≠ []; from specification list%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: y, a1, y1