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Proof step 7 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
s.min ∈ s, s.max ∈ s ⊦ # s = 0, # s = 1, s -- s.min -- (s -- s.min).max = s -- s.max -- (s -- s.max).min, s = ∅, s = ∅ ++ s.min, s = ∅ ++ s.max, s.max < s.min

Rule argument

Additional information

Used simplifier rules: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)