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Proof step 25 from proof sep-ass (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
P0(H1), P1(H2), P2(H), H2 ⊥ H, H1 ⊥ H2, H1 ⊥ H ⊦ (((P0 * P1) * P2))(H1 ∪ H2 ∪ H)

Rule argument

Additional information

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