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Theorem precise-disj

precise(hP0), precise(hP1) ⊦ ((hP0 * hP1 * true))(h0 ∪ h2) ∧ h0 ∪ h2 = h1 ∪ h3 ∧ h0 ⊥ h2 ∧ h1 ⊥ h3 ∧ hP0(h0) ∧ hP1(h1) → h0 ⊥ h1

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