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Proof step 1 from proof size-two-05 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;