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# Re: a problem applying Rule
N Monomachoff wrote:
> Hi there,
> I'm using 3.0.
>
> g[f[f[],a]] /. (n:(f | g))[f[],x__] -> n[f[]] evaluates properly as
> g[f[f[]]].
> But after
> SetAttributes[f,Orderless]
> it no longer works. This doesn't make much sense to me. Is this a bug or
> a feature? And is there a way around this? Thanks for any help,
> nmonomachoff@msn.com
In an earlier posting I wrote
After
SetAttributes[f,Orderless]
The evaluation of
g[f[f[],a]] /. (n:(f | g))[f[],x__] -> n[f[]]
goes through
g[f[a,f[]]] /. (n:(f | g))[f[],x__] -> n[f[]]
hence the result.
But, why , if f is orderless does not (n:(f | g))[f[],x__] imatch the
pattern (n:(f | g))[x__,f[]] ?
Let's look a full forms.
FullForm[(n:(f | g))[f[],x__]]
Pattern[n,Alternatives[f,g]][f[],Pattern[x,BlankSequence[]]]
Presumably the matching works on the basis that
Pattern[n,Alternatives[f,g]] is not an orderless symbol.
If we follow this hint then a cumbersome, way out should be
g[f[f[],a]] /. {f [f[],x__] -> f[f[]], g[f[],x__] -> g[f[]]}
This does work:
SetAttributes[f, Orderless]
g[f[f[],a]] /. {f [f[],x__] -> f[f[]], g[f[],x__] -> g[f[]]}
g[f[f[]]]
g[f[z,f[],a]] /. {f [f[],x__] -> f[f[]], g[f[],x__] -> g[f[]]}
g[f[f[]]]
--
Allan Hayes
Mathematica Training and Consulting
Leicester, UK
hay@haystack.demon.co.uk
http://www.haystack.demon.co.uk
voice: +44 (0)116 271 4198
fax: +44 (0)116 271 8642
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