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

⊦ s ≠ ∅ → ¬ s.max ∈ s.butmax

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmax: ⊦ s ≠ ∅ → s.butmax = s -- s.max;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; from specification set-union