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Proof step 8 from proof sep-partial-mp (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
((P0 * P2 -* P1))(H1), P0(H2), P2(H0), H2 ⊥ H1, H0 ⊥ H2, H0 ⊥ H1 ⊦ P1(H0 ∪ H2 ∪ H1)

Rule argument

Additional information

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