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Re: ReplaceAll / sequential replacements versus joint replacement


Hi Ron,
I agree, it is confusing. But read the Help of ReplaceAll:
"ReplaceAll looks at each part of expr, tries all the rules on it, and 
then goes on to the next part of expr. The first rule that applies to a 
particular part is used; no further rules are tried on that part, or on 
any of its subparts. "

Teherefore, consider e.g.
x /. {x -> a + x, a -> x}
Here only the first rule is applied, giving: a+x

However:
t1 = x /. {x -> a + x}
t1 /. {a -> x}
Here after the first rule did its job, the second rules starts over, 
giving:
2x

good advice -:) read the manual!
Sincerely, Daniel

Ronald Wendner wrote:
> Suppose there is an expression e1 which depends, among others, on variables 
> a and b.
> Replace both a and b by zero:
> 
> e1/.{a->0,b->0} Mathematica gives, say, result1.
> 
> Next, consider
> 
> e2 = e1/.{a->0};
> e2/.{b->0} Mathematica gives, say, result2.
> 
> How can result 1 differ from result 2?
> 
> For my specific problem, I also attach a short Mathematica notebook.
> [Contact the author to obtain the notebook - moderator]
> 
> I'd appreciate any help.
> 
> Many thanks,
> Ron 
> 
> 


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