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Proof step 2 from proof size-three (previous step)

2 < # s ⊦ ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
2 < # s ⊦ ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic