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Proof step 3 from proof indef (previous step)

a ∈ s ⊦ a ∈ s ↔ (∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ s ⊦ ∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0

Rule argument

Additional information

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