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Proof step 11 from proof sep-ass (previous step)

P0(H2), P1(H), P2(H0), H0 ⊥ H, H2 ⊥ H, H ⊥ H0, H0 ⊥ H2 ⊦ ((P0 * P1 * P2))(H2 ∪ H ∪ H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
P0(H2), P1(H), P2(H0), H0 ⊥ H2, H2 ⊥ H, H0 ⊥ H ⊦ ((P0 * P1 * P2))(H2 ∪ H ∪ H0)

Rule argument

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s)