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Proof step 2 from proof one (previous step)

∀ a. a ∈ s → a = b ⊦ s = ∅ ++ b, s = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ s, ∀ a. a ∈ s → a = b ⊦ a0 ∈ s ↔ a0 = b

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%In-empty: ⊦ ¬ a ∈ ∅;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;%ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);