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Proof step 2 from proof min-insert (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 3:
(s ++ a).min < a, a < s.min, s.min ∈ s, s.max ∈ s ⊦ (s ++ a).min = a, s = ∅, # s = 0
Proof step 6:
a < s.min, s.min ∈ s, s.max ∈ s ⊦ (s ++ a).min < a, (s ++ a).min = a, s = ∅, # s = 0

Rule argument

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

Additional information

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