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Proof step 5 from proof bmin-bneq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
s.max = s.min, s.min = (s -- s.min).min, s = ∅ ++ s.min, s.min ∈ s ⊦ # s = 0, # s = 1, s = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ b < a → (a < b ↔ a ≠ b); (ss)% ⊦ ¬ a < a; (s)%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem%in-06: ⊦ a ∈ s ∧ s = ∅ ++ b → a = b; (forward) from specification set-basic% ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; (ws)% ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)