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Proof step 1 from proof k-diff-rec2 (conclusion)

⊦ a ∈ s ∧ a ∈ s0 → s \ s0 = (s -- a) \ (s0 -- a)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: diff-del-minus: ⊦ (s -- a) \ (s1 -- a) = (s -- a) \ s1;%diff-del-minus-3: ⊦ a ∈ s1 → (s -- a) \ s1 = s \ s1;