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Proof step 1 from proof bmin-one (conclusion)

⊦ (∅ ++ a).butmin = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
(∅ ++ a).butmin.min ∈ (∅ ++ a).butmin, (∅ ++ a).butmin.max ∈ (∅ ++ a).butmin ⊦ (∅ ++ a).butmin = ∅, # (∅ ++ a).butmin = 0

Rule argument

Additional information

Used simplifier rules: 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;%minmax-04: ⊦ s ≠ ∅ → ¬ s.max < s.min;%irreflexivity: ⊦ ¬ a < a; from specification oelem%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic