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Proof step 2 from proof minmax-03 (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

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

Rule argument

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

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%seq: ⊦ ¬ b < a → (¬ a < b ↔ a = b); from specification oelem%one-02: ⊦ m ≠ 0 → (¬ 1 < m ↔ m = 1); from specification nat-basic1 Used terms: s