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Proof step 1 from proof minmax-01 (conclusion)

⊦ s.butmin = ∅ ∧ s ≠ ∅ → s.max = s.min

Heuristic simplifier applied rule simplifier

And closed the goal!

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;%min-one-01: ⊦ # s = 1 → ∅ ++ s.min = s;%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%one-in: ⊦ # s = 1 ∧ a ∈ s → s = ∅ ++ a; (forward) from specification set-basic%size-one-02: ⊦ # s = 1 ∧ a ∈ s ∧ b ∈ s → a = b; (forward) from specification set-basic%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