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

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

User applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: (s ++ a).min ∈ s

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min; Used formulas: (s ++ a).min ∈ s