Re: [stack] function/object ambiguity + quotation alternative

On Feb 26, 2009, at 12:22 PM, Stevan Apter wrote:

>>   2. [dup] == [dup] first      (removing shared 'i' suffix)
>
> no.  dup = [dup] first.

You're missing the point. I know that 'dup = [dup] first', but I can  
prove '[dup] == [dup] first'!

I'll start with your example (minus the swap); after each step, I'll  
indicate how I came upon the given equality:

1. 2 [dup] first i == 2 2        (given)
2. [F] i == F                    (given)
3  2 dup == 2 2                  (given)
4. 2 [dup] i == 2 2              (2+3: substituted '[dup] i' for 'dup')
5. 2 [dup] i == 2 [dup] first i  (1+4: both equal '2 2')
6. i == first i                  (5: removed shared '2 [dup]' prefix)
7. id == first                   (6: removed 'i' suffix)

To repeat, I know this is a bogus conclusion; that's my point. The  
problem arises from the fact that '[dup] i' functions identically to  
'[dup] first i' in XY; or, at least that's my understanding.

If my little derivation is wrong, please correct me. If not, I believe  
this to be a problem in XY's semantics.

- John