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From: Wolfgang Lux <wlux_at_uni-muenster.de>

Date: Fri, 09 Nov 2007 17:10:00 +0100

Hi Sergio!

*> OK, but now consider adding
*

*>
*

*> f False = failed
*

*>
*

*> The meaning of the program should remain the same, but the step
*

*> in question becomes much more plausible.
*

Thank you for this clarification. So the problem can be boiled

down to this one. Given the functions

f False = 1 + failed

g False = success

there is a reduction of

const (f x) (g x)

to success even if g is rigid, namely with the following steps

const (f x) (g x)

~~> {x=False} const (1 + failed) (g False) -- narrowing (f x)

~~> {x=False} const (1 + failed) success -- reducing (g False)

~~> {x=False} succeess -- reducing const ...

and this result cannot be computed with either an outer-most strategy

(assuming g is rigid) nor an inner-most strategy (because 1+failed

has no solution).

Regards

Wolfgang

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Received on Fr Nov 09 2007 - 17:51:51 CET

Date: Fri, 09 Nov 2007 17:10:00 +0100

Hi Sergio!

Thank you for this clarification. So the problem can be boiled

down to this one. Given the functions

f False = 1 + failed

g False = success

there is a reduction of

const (f x) (g x)

to success even if g is rigid, namely with the following steps

const (f x) (g x)

~~> {x=False} const (1 + failed) (g False) -- narrowing (f x)

~~> {x=False} const (1 + failed) success -- reducing (g False)

~~> {x=False} succeess -- reducing const ...

and this result cannot be computed with either an outer-most strategy

(assuming g is rigid) nor an inner-most strategy (because 1+failed

has no solution).

Regards

Wolfgang

_______________________________________________

curry mailing list

curry_at_lists.RWTH-Aachen.DE

http://MailMan.RWTH-Aachen.DE/mailman/listinfo/curry

Received on Fr Nov 09 2007 - 17:51:51 CET

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