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

⊦ s1 \ (s2 ++ a) = (s1 \ s2) -- a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ a0 ∈ s1 \ (s2 ++ a) ↔ a0 ≠ a ∧ a0 ∈ s1 \ s2

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)