Re: ReplaceAll / sequential replacements versus joint replacement

• To: mathgroup at smc.vnet.net
• Subject: [mg55449] Re: ReplaceAll / sequential replacements versus joint replacement
• From: dh <dh at metrohm.ch>
• Date: Thu, 24 Mar 2005 03:41:49 -0500 (EST)
• References: <d1rhch\$45j\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```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

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
>
>

```

• Prev by Date: Hypergeometric integral looks wrong ?
• Next by Date: Re: ReplaceAll / sequential replacements versus joint replacement
• Previous by thread: Re: ReplaceAll / sequential replacements versus joint replacement
• Next by thread: Re: ReplaceAll / sequential replacements versus joint replacement