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Proof step 2 from proof bmin-in-01 (previous step)

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;%Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;% ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; (ws)% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic