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Proof step 3 from proof sub-one-01 (previous step)

∀ a0. a0 ∈ s → a0 = a ⊦ ∀ a0. ¬ a0 ∈ s, a0 ∈ s ↔ a0 = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a1 ∈ s, ∀ a0. a0 ∈ s → a0 = a ⊦ a0 ∈ s ↔ a0 = a

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);