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Proof step 4 from proof sep-partial-mp-2 (previous step)

⊦ ∀ P0. ((P * P0))(H) → (∃ H0, H1. H0 ⊥ H1 ∧ H0 ∪ H1 = H ∧ P(H0) ∧ P0(H1))

Lemma sep-split-interactive