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Proof step 24 from proof sharing-com (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
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)