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Proof step 2 from proof prefix-ex (previous step)

x ⊑ y + z ⊦ x ⊑ y, ∃ x0. x = y + x0 ∧ x0 ⊑ z

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ 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

Rule argument

indvar: x

Additional information

Used simplifier rules: prefix-nil: ⊦ [] ⊑ x;%list freely generated by [], +;;