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Proof step 2 from proof k-union-rec2 (previous step)

⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a, a ∈ s0, a ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: unadd: ⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)