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

# s = 2, a ∈ s ⊦ ∃ 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 = 2, a ∈ s ⊦ ∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b

Rule argument

fma: a ∈ s

Additional information

Used formulas: a ∈ s