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Proof step 1 from proof minmax (conclusion)

⊦ s.butmin = ∅ ∧ s ≠ ∅ → s.butmax = ∅

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%Butmax: ⊦ s ≠ ∅ → s.butmax = s -- s.max;%Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;%min-one-01: ⊦ # s = 1 → ∅ ++ s.min = s;%max-one-01: ⊦ # s = 1 → (s = ∅ ++ s.max ↔ true);%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%one-size: ⊦ s = ∅ ++ a → # s = 1; (forward) from specification set-basic%del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union