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Proof step 8 from proof sep-ass (previous step)

P2(H0), H1 ⊥ H0, ∃ H0, H2. H0 ⊥ H2 ∧ H1 = H0 ∪ H2 ∧ P0(H0) ∧ P1(H2) ⊦ ((P0 * P1 * P2))(H1 ∪ H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
P0(H2), P1(H), P2(H0), H2 ⊥ H, H0 ⊥ H2, H0 ⊥ H ⊦ ((P0 * P1 * P2))((H2 ∪ H) ∪ H0)

Rule argument

Additional information

Used simplifier rules: ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2); (ws)