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Proof step 1 from proof uni-ass (conclusion)

⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H

System applied rule normalize

And obtained the following new goals

Proof step 2:
H0 ⊥ H1, H1 ⊥ H, H ⊥ H0 ⊦ (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H

Rule argument

Additional information