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Proof step 1 from proof bmin-in-03 (conclusion)

⊦ a ∈ s.butmin ∧ s ≠ ∅ ↔ a ∈ s ∧ a ≠ 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;% ⊦ ¬ a ∈ ∅; (s)% ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; (ws)