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Proof step 3 from proof union-diff-union-same (previous step)

⊦ ∀ a. a ∈ (s0 ∪ s1) \ (s2 ∪ s1) ↔ a ∈ s0 \ (s2 ∪ s1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ a ∈ (s0 ∪ s1) \ (s2 ∪ s1) ↔ a ∈ s0 \ (s2 ∪ s1)

Rule argument

Additional information