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Proof step 8 from proof list-split (previous step)

# x ≤ # y0 + 1, x.tail ⊑ y0 + z, ∀ x, z. # x ≤ # y0 ∧ x ⊑ y0 + z → (∃ y1. y0 = x + y1) ⊦ x = [], ∃ y. x.head ' + y0 = x + y

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
x = a + y2, # x ≤ # y0 + 1, y2 ⊑ y0 + z, ∀ x, z. # x ≤ # y0 ∧ x ⊑ y0 + z → (∃ y1. y0 = x + y1) ⊦ x = [], ∃ y. a ' + y0 = x + y

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x, a, y2]

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x, a, y2