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Proof step 2 from proof uni-ass (previous step)

H0 ⊥ H1, H1 ⊥ H, H ⊥ H0 ⊦ (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
H ⊥ H0, H1 ⊥ H, H0 ⊥ H1 ⊦ (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H

Rule argument

Additional information