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Proof step 3 from proof prefix-ex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ' + x0 ⊑ y + z, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ a ' + x0 ⊑ y, ∃ x. a ' + x0 = y + x ∧ x ⊑ z

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list