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Proof step 1 from proof diff-del (conclusion)

⊦ s = s1 \ s2 ∧ ¬ a ∈ s → (s = (s1 -- a) \ (s2 -- a) ↔ true)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ s1 \ s2 = (s1 -- a) \ s2, a ∈ s1 \ s2

Rule argument

Additional information

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