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Proof step 1 from proof diff-rc (conclusion)

⊦ (s \ s1) \ s2 = (s \ s2) \ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ a ∈ (s \ s1) \ s2 ↔ a ∈ (s \ s2) \ s1

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic