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Proof step 1 from proof bmin-in-01 (conclusion)

⊦ a ∈ s ∧ a ≠ s.min → a ∈ s.butmin

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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