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Proof step 2 from proof int-ins (previous step)

⊦ (∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0) ↔ (∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s

User applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0 ⊦ (∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s
Proof step 9:
(∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s ⊦ ∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0

Rule argument

pos: right 1

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