Terrence, what do you mean by 

>  Both (i.4) +/ (i.4)  and i.4 +/ (i.4) parse the same

Do  3+4 * (6+7)  and  (3+4) * (6+7)  parse the same too?  Nope.

In  (i.4) +/ (i.4)  you've got:

           y =. i.4
           x =. i.4
           
           x +/ y
        
   
in   i.4  +/ (i.4)  you've got:

           y =. i.4
           x =.   4

           z =. x +/ y
           
           i. z
Try:

           load 'trace'
        
           trace '(i.4) +/ (i.4)'
        
           trace ' i.4  +/ (i.4)'
        
-Dan
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