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Proof step 1 from proof diff-del-minus-3 (conclusion)

⊦ a ∈ s1 → (s -- a) \ s1 = s \ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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