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Theorem precise-disj

precise(P0), precise(P1) ⊦ ((P0 * P1 * true))(H0 ∪ H2) ∧ H0 ∪ H2 = H1 ∪ H3 ∧ H0 ⊥ H2 ∧ H1 ⊥ H3 ∧ P0(H0) ∧ P1(H1) → H0 ⊥ H1

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