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

ordered≤(x0), x1 ⊑ x0.rest, (∃ x0. x1 ⊑ x0 ∧ ordered≤(x0)) → ordered≤(x1) ⊦ x0 = [], ordered≤(x0.first ' + x1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
x0 = a + y, ordered≤(x0), x1 ⊑ y, (∃ x0. x1 ⊑ x0 ∧ ordered≤(x0)) → ordered≤(x1) ⊦ x0 = [], ordered≤(a ' + x1)

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] → [x0, a, y] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x0, a, y