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Proof step 3 from proof minmax-03 (previous step)

s.max = s.min, s.min ∈ s ⊦ # s = 0, # s = 1, s = ∅

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
1 < # s ⊦ ∃ a, b. a ≠ b ∧ a ∈ s ∧ b ∈ s

Rule argument

spec: set-basic name: size-two-03 seq: 1 < # s ⊦ ∃ a, b. a ≠ b ∧ a ∈ s ∧ b ∈ s subst: [s] → [s]

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem% ⊦ ¬ a < a; (s)% ⊦ ¬ b < a → (a < b ↔ a ≠ b); (ss)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) Used terms: s