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Proof step 7 from proof bm-shift (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union