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Proof step 1 from proof undel-02 (conclusion)

⊦ (s ∪ s0) -- a = (s -- a) ∪ (s0 -- a)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: undel-04: ⊦ (s -- a) ∪ (s0 -- a) = (s ∪ s0) -- a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;