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Proof step 4 from proof butmin-ins (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
s.min < a, s.min ∈ s, s.max ∈ s ⊦ s = ∅, (s ++ a) -- (s ++ a).min = (s -- s.min) ++ a, # s = 0

Rule argument

Additional information

Used simplifier rules: ne: ⊦ ∅ ≠ s ++ a; from specification set-basic