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Proof step 7 from proof size-three (previous step)

∃ b. b ∈ s ∧ b ≠ a, 2 < # s, a ∈ s ⊦ # s = 0, ∃ 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

Rule argument

Additional information