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

s.butmin.min < s.min ⊦ s.butmin = ∅, s = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
(s -- s.min).min < s.min, s.min ∈ s, s.max ∈ s ⊦ # s = 0, s = ∅, s = ∅ ++ s.min, 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)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic% ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)