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Proof step 5 from proof size-three-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# s = 3, a ∈ s ⊦ ∃ a, b, c. a ≠ b ∧ a ≠ c ∧ b ≠ c ∧ s = ∅ ++ a ++ b ++ c

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);