> left rank of $ is 1 so that the scalar 0 in (0 $ 2) will be 
> extended into ,0 
> (rank-1) by J before execution.

Hold it - that's not right.  Left rank of 1 means that higher
ranks are broken into 1-cells, but it DOES NOT mean that
scalars are turned into 1-cells before the verb operates on
them.  The verb has to be prepared to deal with scalars.
If it wants to convert them to 1-cells, it may, but it doesn't
have to.

Henry Rich

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