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

⊦ a0 ∈ s ↔ a0 = a, (∀ a1. a1 ∈ s → a1 = a) ↔ (∀ a1. ¬ a1 ∈ s)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 4:
∀ a1. a1 ∈ s → a1 = a ⊦ a0 ∈ s ↔ a0 = a, ∀ a1. ¬ a1 ∈ s
Proof step 8:
∀ a1. ¬ a1 ∈ s ⊦ a0 ∈ s ↔ a0 = a, ∀ a1. a1 ∈ s → a1 = a

Rule argument

pos: right 2

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