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Proof step 10 from proof prefix-ex (previous step)

y = a + y0, x0 ⊑ y0 + z, ∀ 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 11:
x0 ⊑ y0 + z, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ x0 ⊑ y0, ∃ x. x0 = y0 + x ∧ x ⊑ z

Rule argument

Additional information

Used simplifier rules: postifx-app-03: ⊦ x + y ⊑ x + z ↔ y ⊑ z;% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)