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Proof step 1 from proof prefix-ex (conclusion)

⊦ ¬ x ⊑ y → (x ⊑ y + z ↔ (∃ x0. x = y + x0 ∧ x0 ⊑ z))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x ⊑ y + z ⊦ x ⊑ y, ∃ x0. x = y + x0 ∧ x0 ⊑ z

Rule argument

Additional information

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