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Proof step 6 from proof bmin-bin (previous step)

s.max = s.min, s = ∅ ++ s.min, s.min ∈ s ⊦ # s = 0, # s = 1, s = ∅, (s -- s.min).min ∈ s

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 2 flags: discard

Additional information

Used simplifier rules: max-one: ⊦ (∅ ++ a).max = a;%min-one: ⊦ (∅ ++ a).min = a;% ⊦ ∅ ≠ s ++ a; (s)% ⊦ #(∅ ++ a) = 1; (s)