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Proof step 2 from proof del-e-02 (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ (∀ a0. ¬ a0 ∈ s) ∨ (∀ a0. a0 ∈ s ↔ a0 = a)
Proof step 9:
(∀ a0. ¬ a0 ∈ s) ∨ (∀ a0. a0 ∈ s ↔ a0 = a) ⊦ ∀ a0. a0 = a ∨ ¬ a0 ∈ s

Rule argument

pos: right 1

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