Hello.

> 2. 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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