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Proof step 22 from proof btree-clean-fwd (previous step)

⊦ ((P * P2))(zh) ↔ (∃ zh0, zh1. zh0 ⊥ zh1 ∧ zh = zh0 ∪ zh1 ∧ P(zh0) ∧ P2(zh1))

A spec-lemma