We could derive set equality from Equal[A]
, but it would be O(n^2)
.
Instead, we require
Order[A], reducing the complexity to
O(log n)
We could derive set equality from Equal[A]
, but it would be O(n^2)
.
Instead, we require
Order[A], reducing the complexity to
O(log n)
If Equal[A].equalIsNatural == true
, than Any#==
is used.