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Proof step 6 from proof size-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ∈ s ⊦ # s = 2 ↔ (∃ 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);