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


From: William Sit
Subject: Re: [Axiom-developer] "has" and "with" (was curious algebra failure)
Date: Sun, 12 Aug 2007 21:40:53 -0400

On Sun, 12 Aug 2007 20:19:47 -0500 (CDT)
 Gabriel Dos Reis <address@hidden> wrote:
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

Isn't the problem whether 'Monad has SetCategory with "*":(%,%)->%' ? My answer would be yes. This should not be related to equivalence of categories or domains. We already have seen examples that Integer has with _*:(Integer, Integer)->Integer. Here, we are not saying 'Monad is SetCategory with "*":(%,%)->%'.

William




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