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Proof step 2 from proof intu-empty (previous step)

⊦ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∪ s1) ↔ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0) ∧ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∪ s1 ⊦ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0) ∧ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1)
Proof step 9:
∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1 ⊦ ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∪ s1

Rule argument

pos: right 1

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