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Proof step 1 from proof size-one-02 (conclusion)

⊦ # s = 1 ∧ a ∈ s ∧ b ∈ s → a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# s = 1, b ∈ s, a ∈ s ⊦ a = b

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);