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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:
⊦ a0 ∈ s1 \ s2 ↔ a0 ∈ (s1 -- a) \ (s2 -- a), a ∈ s1 \ s2

Rule argument

Additional information

Used simplifier rules: ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);