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Proof step 7 from proof true-elims (
previous step
)
⊦ ((P0 ⊗ P1))(H) ↔ (∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P1(H1 ∪ H2))
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