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Hello.
- Designers writing the simplest clearest code they can for operations
but not thinking about all the boundary cases. (x-x and x+x are perfectly reasonable in arithmetic; x\x and xUx are perfectly reasonable in set theory; but they are a sort of boundary case.)
x\x and xUx are not equivalent to removeAll: and addAll: because \ and U don't mutate the contents of x.
The operation \ would correspond to copyWithoutAll:, while U would correspond to copyWithAll:.
Andres.
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