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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# s = 2, a ∈ s ⊦ ∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b

Rule argument

Additional information