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

1 < # s ⊦ s.min < s.max

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
s.max = s.min, s.min ∈ s ⊦ # s = 0, # s = 1, s = ∅

Rule argument

Additional information

Used simplifier rules: Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem% ⊦ ¬ b < a → (a < b ↔ a ≠ b); (ss)% ⊦ ¬ a < a; (s)%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)