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Proof step 2 from proof sub-one-01 (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a0. a0 ∈ s → a0 = a ⊦ ∀ a0. ¬ a0 ∈ s, a0 ∈ s ↔ a0 = a
Proof step 7:
∀ a0. ¬ a0 ∈ s ⊦ ∀ a0. a0 ∈ s → a0 = a, a0 ∈ s ↔ a0 = a

Rule argument

pos: right 1

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