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

Rule argument

pos: right 1

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