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Proof step 10 from proof size-two (previous step)

⊦ 1 = # s0 ↔ (∃ a0, b. a0 ≠ b ∧ s0 ++ a = ∅ ++ a0 ++ b), a ∈ s0

User applied rule case distinction

And obtained the following new goals

Proof step 11:
1 = # s0 ⊦ a ∈ s0, ∃ a0, b. a0 ≠ b ∧ s0 ++ a = ∅ ++ a0 ++ b
Proof step 15:
∃ a0, b. a0 ≠ b ∧ s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a ∈ s0

Rule argument

pos: right 1

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