Re: [Axiom-developer] "has" and "with" (was curious algebra failure)

[Gabriel Dos Reis]

On Sun, 12 Aug 2007, Bill Page wrote:

| On 8/12/07, Gabriel Dos Reis wrote:
| > On Sun, 12 Aug 2007, Bill Page wrote:
| > | ...
| > | It should be fine because of rule 2:
| > |
| > | 2. Anonymous types are equivalent when structurally equivalent
| >
| > The parameter S in RepeatedSquare(S) of the category
| >
| >    SetCategory with "*": (%,%) -> %
| >
| > but RepeatedSquare is being called with a domain of a named category 
(Monad).
| > Rule 2 says:
| >
| >   2. Anonymous types are equivalent when stucturally equivalent
| >
| > How would it apply?
| >
| 
| Why do you think:
| 
|      import RepeatedSquaring(%)
| 
| is referring to a named category?

I'm not saying that.  

I'm saying that the parameter S of the default package Monad& -- generated for
the default implementation of the category Monad -- is of the named category
Monad.  It is that parameter S which is  being used to instantiate
RepeatedSquaring.  However, RepeatedSquaring expects its (domain) argument to
be of the unnamed category 

   SetCategory with "*": (%,%) -> %

-- Gaby