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Proof step 8 from proof min-delmin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
s.max = s.min, s = ∅ ++ s.min, s.min ∈ s ⊦ # s = 1, # s = 0, s = ∅, s.min < (s -- s.min).min, (s -- s.min).min ≠ s.min ∧ (s -- s.min).min ∈ s

Rule argument

Additional information

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