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Proof step 1 from proof unadd (conclusion)

⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: unadd-01: ⊦ s ∪ (s0 ++ a) = (s ∪ s0) ++ a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;