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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# s = 2, a ∈ s ⊦ ∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b

Rule argument

Additional information

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