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

# s ≠ 0 → (∃ a. a ∈ s), 2 < # s ⊦ # s = 0, s = ∅, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 < # s, a ∈ s ⊦ # s = 0, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Rule argument

Additional information

Used simplifier rules: ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);