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Proof step 2 from proof bmin-one-02 (previous step)

s.butmin = ∅ ⊦ a ∈ s ↔ a = s.min ∧ s ≠ ∅

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%min-less: ⊦ a ≠ s.min ∧ a ∈ s → s.min < a;%Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;%Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%del-in: ⊦ ¬ a ∈ s → s -- a = s; from specification set-union%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union%sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%one-size: ⊦ s = ∅ ++ a → # s = 1; (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