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Proof step 3 from proof eq-01 (previous step)

⊦ (∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)) ↔ (∀ a. a ∈ s ↔ a ∈ s0), a ∈ s, a ∈ s0

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 4:
∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a ∈ s0, a ∈ s, ∀ a. a ∈ s ↔ a ∈ s0
Proof step 8:
∀ a. a ∈ s ↔ a ∈ s0 ⊦ a ∈ s0, a ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)

Rule argument

pos: right 1

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