[stack] Re: unary functions from X to Y?

--- In [email protected], Chris Double <chris.double@...> wrote:
>
> On Fri, Mar 6, 2009 at 7:30 PM, Justin Pombrio <zallambo@...> wrote:
> > For instance, suppose 'two!' takes any stack and produces a stack with just
> > one element, '2'. Then 'f two!' should produce the stack '2' for all terms
> > f. But:
> >
> > 3 abort two! == 3
> > 3 => two! == 2 3.
> >
> 
> I'm not sure this is correct for XY. 'f two!' for XY, where f is any
> term,  will always produce a stack '2'. The example you give is
> effectively:
> 
> > two! 3 == 2 3
> 
> This is because the '3' is appended to the tail of the queue. Which
> then gets executed. The function 'two!' is still give a stack (which
> is empty) and returns a stack containing 2.
> 
> Chris.
> -- 
> http://www.bluishcoder.co.nz
>

Do you agree that the following two equivalences hold?
3 => two!   ==   two! 3   ==   2 3

I'm not saying that this is nonsensical or a contradiction, it's just not how *function composition* works. If this were actually function composition, then *any* program ending with the term 'two!' would produce the stack '2'.

I should have been more specific. I meant that 'P two!' should always produce the stack '2' for *any* program P.

Justin