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Proof step 10 from proof bm-shift (previous step)

s.max ∈ s, s.min ∈ s ⊦ # s = 1, # s = 0, s -- s.min -- (s -- s.min).max = s -- s.max -- (s -- s.max).min, s = ∅ ++ s.max, s = ∅ ++ s.min, s = ∅, s.max < s.min, s.min ≠ s.max ∧ s ≠ ∅

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: minmax-02: ⊦ s.min = s.max ∧ s ≠ ∅ → # s = 1;