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Proof step 27 from proof sharing-com (previous step)

P0(H1 ∪ H0), P1(H2 ∪ H0), H1 ⊥ H0, H2 ⊥ H1, H1 ⊥ H2, H2 ⊥ H0, H0 ⊥ H1 ⊦ H2 ∪ H1 ∪ H0 = (H1 ∪ H2) ∪ H0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
P0(H1 ∪ H0), P1(H2 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0 ⊦ H2 ∪ H1 ∪ H0 = (H1 ∪ H2) ∪ H0

Rule argument

Additional information

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