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Proof step 2 from proof minmax-02 (previous step)

s.min = s.max, s.max ∈ s ⊦ # s = 1, s = ∅, # s = 0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
s.min = s.max, s.max ∈ s ⊦ s.min < s.max, # s = 1, s = ∅, # s = 0

Rule argument

fma: s.min < s.max

Additional information

Used simplifier rules: max-gr: ⊦ s.max < a ∧ s ≠ ∅ → ¬ a ∈ s;%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem Used formulas: s.min < s.max