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Proof step 2 from proof max-bmin (previous step)

1 < # s ⊦ s.butmin.max = s.max

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%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;% ⊦ ¬ 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)