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Proof step 3 from proof min-insert (previous step)

a < 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 4:
a < s.min, (s ++ a).min < a, s.max ∈ s, s.min ∈ s ⊦ # s = 0, (s ++ a).min = a, s = ∅
Proof step 7:
a < s.min, s.max ∈ s, s.min ∈ s ⊦ # s = 0, (s ++ a).min = a, s = ∅, (s ++ a).min < a

Rule argument

fma: (s ++ a).min < a

Additional information

Used formulas: (s ++ a).min < a