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Proof step 3 from proof postfix-split-stronger (previous step)

x ⊒ y + z, ∀ x0, y, z. # x0 < # x → x0 ⊒ y + z → x0 ⊒ z ∨ (∃ x1. x1 + z = x0 ∧ x1 ⊒ y) ⊦ x ⊒ z ∨ (∃ x0. x0 + z = x ∧ x0 ⊒ y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information