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AW: Strange ReplaceAll behavior


Hello Lawrence,

using Set (=) instead of SetDelayed (:=) in 

f[x_]:=(a+b) /. ru[a]; 

will give you what you desire.

Best regards,

Matthias Bode.

-----Ursprüngliche Nachricht-----
Von: Lawrence A. Walker Jr. [mailto:lwalker701 at earthlink.net]
Gesendet: Mittwoch, 25. September 2002 07:51
An: mathgroup at smc.vnet.net
Betreff: [mg36774] Strange ReplaceAll behavior


Hi,

For the life of me I am not sure why the following is not working in my 
v. 4.2:

ru[a]=a->x;
f[x_]:=(a+b) /. ru[a];

Why do I get
f[c] = b+x

and not
f[c] = b+c?

What gives?

Thanks,
Lawrence

-- 
Lawrence A. Walker Jr.
http://www.kingshonor.com


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