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Proof step 2 from proof diff-rc (previous step)

⊦ a ∈ (s \ s1) \ s2 ↔ a ∈ (s \ s2) \ s1

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;