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Proof step 1 from proof diff-del-minus-2 (conclusion)

⊦ ¬ a ∈ s1 → (s -- a) \ s1 = (s \ s1) -- a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ a0 ∈ (s -- a) \ s1 ↔ a0 ∈ (s \ s1) -- a, a ∈ s1

Rule argument

Additional information

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