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Theorem sep

⊦ (R0 * R1)(H0, H1) ↔ (∃ H2, H3, H4, H5. H2 ⊥ H3 ∧ H4 ⊥ H5 ∧ H2 ∪ H3 = H0 ∧ H4 ∪ H5 = H1 ∧ R0(H2, H4) ∧ R1(H3, H5))