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Proof step 4 from proof prefix-trans (previous step)

x ⊑ y, y ⊑ a ' + z0, x.rest ⊑ y.rest ∧ y.rest ⊑ z0 → x.rest ⊑ z0, ∀ x, y. x ⊑ y ∧ y ⊑ z0 → x ⊑ z0 ⊦ x ⊑ a ' + z0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
x = a0 + y0 ∧ x ≠ [], x ⊑ y, y ⊑ a ' + z0, y0 ⊑ y.rest ∧ y.rest ⊑ z0 → y0 ⊑ z0, ∀ x, y. x ⊑ y ∧ y ⊑ z0 → x ⊑ z0 ⊦ x ⊑ a ' + z0

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] → [x, a0, y0] flags:

Additional information

Used simplifier rules: prefix-nil: ⊦ [] ⊑ x;%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a0, y0