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Proof step 5 from proof precise-sep (previous step)

precise(P0), precise(P1) ⊦ ∀ H, H0, H1, H2. ((P0 * P1))(H) ∧ ((P0 * P1))(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information