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

⊦ (s -- a) \ (s1 -- a) = (s -- a) \ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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