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

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 3 flags: discard

Additional information

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