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Proof step 1 from proof two-01 (conclusion)

a ∈ s, s ≠ ∅ ++ a ⊦ ∃ b. b ∈ s ∧ b ≠ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ s, a ∈ s ⊦ a0 = a, ∃ b. b ∈ s ∧ b ≠ a

Rule argument

Additional information

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