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Proof step 9 from proof del-ne (previous step)

a1 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s, b = a ∨ ¬ b ∈ s ⊦ # s = 0, # s = 1, a1 = b

User applied rule case distinction

And obtained the following new goals

Proof step 10:
b = a, a ∈ s, b ∈ s, a1 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ # s = 1, # s = 0, a1 = b

Rule argument

pos: left 5

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