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Proof step 5 from proof min-insert (previous step)

(s ++ a).min < a, a < s.min, s.min ∈ s, s.max ∈ s, (s ++ a).min < s.min ⊦ (s ++ a).min ∈ s, (s ++ a).min = a, s = ∅, # s = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: (s ++ a).min ∈ s ++ a

Additional information

Used simplifier rules: min-inn: ⊦ ¬ s.min ∈ s ↔ s = ∅;%ne: ⊦ ∅ ≠ s ++ a; from specification set-basic%in-02: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s); from specification set-basic Used formulas: (s ++ a).min ∈ s ++ a