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Proof step 1 from proof size-two-04 (conclusion)

⊦ # s ≠ 0 ∧ # s ≠ 1 ∧ a ∈ s → (∃ b. b ∈ s ∧ a ≠ b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s ⊦ # s = 0, # s = 1, ∃ b. b ∈ s ∧ a ≠ b

Rule argument

Additional information

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