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Proof step 1 from proof bmin-two-06 (conclusion)

⊦ # s = 2 → s.butmin.min ∈ s.butmin

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: min-inn: ⊦ ¬ s.min ∈ s ↔ s = ∅;%bmin-two-07: ⊦ # s = 2 → s.butmin ≠ ∅;