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Proof step 8 from proof butmin-ins (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
s.min < a, s.min ∈ s, s.max ∈ s ⊦ # s = 0, (s ++ a) -- s.min = (s -- s.min) ++ a, s = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ a < b → (b < a ↔ false); (ss)