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Proof step 2 from proof eq-01 (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ ∀ a. a ∈ s ↔ a ∈ s0, a ∈ s0, a ∈ s
Proof step 7:
∀ a. a ∈ s ↔ a ∈ s0 ⊦ ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0), a ∈ s0, a ∈ s

Rule argument

pos: right 1

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