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Proof step 2 from proof diff-del-plus-2 (previous step)

⊦ (s ++ a) \ s1 = (s \ s1) ++ a, a ∈ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ a0 ∈ (s ++ a) \ s1 ↔ a0 = a ∨ a0 ∈ s \ s1, a ∈ s1

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)