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Proof step 5 from proof size-two (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a ∈ s, # s ≠ 0 → a ∈ s ⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b)

Rule argument

fma: a ∈ s

Additional information

Used simplifier rules: size-zero-02: ⊦ # s = 0 → s = ∅;%Size-empty: ⊦ # ∅ = 0;%ne: ⊦ ∅ ≠ s ++ a; Used formulas: a ∈ s