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Proof step 19 from proof precise-sep (
previous step
)
⊦ precise(P) ↔ (∀ H, H0, H1, H2. P(H) ∧ P(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0)
Lemma
precise