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Proof step 8 from proof prefix-ex (previous step)

a = (y + z).head, x0 ⊑ (y + z).tail, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ y + z = [], y = [], a ' + x0 ⊑ y, ∃ x. a ' + x0 = y + x ∧ x ⊑ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
x0 ⊑ y.tail + z, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ y + z = [], y = [], y.head ' + x0 ⊑ y, ∃ x. y.head ' + x0 = y + x ∧ x ⊑ z

Rule argument

Additional information

Used simplifier rules: ⊦ x ≠ [] → (x + y).head = x.head; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x ≠ [] → (x + y).tail = x.tail + y; (s)