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Proof step 2 from proof bmin-one-01 (previous step)

s.butmin = ∅ ⊦ # s = 1, s = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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;%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem% ⊦ ¬ a < a; (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%in-06: ⊦ a ∈ s ∧ s = ∅ ++ b → a = b; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic% ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)