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Proof step 2 from proof in-03 (previous step)

⊦ a ∈ s ++ b ↔ a = b, a ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in-02: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s);%add: ⊦ a ∈ s ++ b ↔ ¬(a ≠ b ∧ ¬ a ∈ s);