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Proof step 10 from proof intu-empty (previous step)

a ∈ s, a ∈ s0 ∨ a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ⊦ s = ∅, # s = 0

User applied rule case distinction

And obtained the following new goals

Proof step 11:
a ∈ s0, a ∈ s, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ⊦ s = ∅, # s = 0
Proof step 13:
a ∈ s1, a ∈ s, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ⊦ s = ∅, # s = 0

Rule argument

pos: left 2

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