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Proof step 4 from proof size-three-02 (previous step)

# s ≠ 0 → a ∈ s ⊦ # s = 3 ↔ (∃ a, b, c. a ≠ b ∧ a ≠ c ∧ b ≠ c ∧ s = ∅ ++ a ++ b ++ c)

User applied rule case distinction

And obtained the following new goals

Proof step 5:
# s = 3, # s ≠ 0 → a ∈ s ⊦ ∃ a, b, c. a ≠ b ∧ a ≠ c ∧ b ≠ c ∧ s = ∅ ++ a ++ b ++ c
Proof step 21:
∃ a, b, c. a ≠ b ∧ a ≠ c ∧ b ≠ c ∧ s = ∅ ++ a ++ b ++ c, # s ≠ 0 → a ∈ s ⊦ # s = 3

Rule argument

pos: right 1

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