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Proof step 41 from proof uni-ass (conclusion)

⊦ 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 40:
H ⊥ H0, H1 ⊥ H, H0 ⊥ H1 ⊦ (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]