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Proof step 6 from proof min-insert (previous step)

(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

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 ≠ ∅;% ⊦ ∅ ≠ s ++ a; (s)% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (ws) Used formulas: (s ++ a).min ∈ s ++ a