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Proof step 1 from proof prefix-split-stronger (conclusion)

x ⊑ y + z, ¬ x ⊑ y ⊦ ∃ 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