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Proof step 4 from proof intu-empty (previous step)

∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∪ s1 ⊦ (¬ a ∈ s ∨ ¬ a ∈ s0) ∧ (¬ a0 ∈ s ∨ ¬ a0 ∈ s1)

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∪ s1 ⊦ ¬ a ∈ s ∨ ¬ a ∈ s0
Proof step 7:
∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∪ s1 ⊦ ¬ a0 ∈ s ∨ ¬ a0 ∈ s1

Rule argument

pos: right 1

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