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I am trying to model certain non-commutative, but associative, structures with Mathematica. As a notation for the product I take CenterDot and add the attribute Flat since the product is associative. Then, I would like to realize certain relations, e.g. $$ a \cdot f[b,c]=f[a\cdot b,c]+f[a,b\cdot c] $$ for a function $f[b,c]$. I thought that this can be done systematically with a simple replace rules

$$a \cdot f[b,c]\,/. a\_ \cdot f[b\_,c\_]\rightarrow f[a\cdot b,c]+f[a,b\cdot c]$$

However, the output of such an operation is (the second term is ok)

$$f[\text{CenterDot}[a],b\cdot c]+f[a\cdot b,c]$$

If I use BlankSequence, $a\_\_$ instead of Blank, the output is exactly I want, but then the rule fails to give the right answer when there are several factors on the left (ideally, I would use the rule as many times as to remove all factors from the left. For example,

$$u\cdot a \cdot f[b,c]$$

is transformed to (in the second term it generates extra arguments, which is meaningless) $$f[u\cdot a\cdot b,c]+f[u,a,b\cdot c]$$

Ironically, the first rule (with Blank) works here and gives the desired

$$f[u\cdot a,b\cdot c]+f[u\cdot a\cdot b,c]$$

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The OneIdentity attribute will help:

SetAttributes[CenterDot, {Flat, OneIdentity}];

a_ \[CenterDot] f[b_,c_] := f[a \[CenterDot] b, c] + f[a, b \[CenterDot] c]

Then:

a \[CenterDot] f[b, c] //TeXForm

$f(a,b\cdot c)+f(a\cdot b,c)$

And your second example:

CenterDot[u, a, f[b,c]] //TeXForm

$f(u\cdot a,b\cdot c)+f(u\cdot a\cdot b,c)$

Addendum

Why does OneIdentity help?

When pattern matching, note the way Flat works:

ClearAll[CenterDot]
SetAttributes[CenterDot, Flat]

CenterDot[x, y] /. CenterDot[a_, b_] :> {a,b}

{CenterDot[x], CenterDot[y]}

The arguments a and b have acquired the head CenterDot.

(This does not always happen. Here are a few examples where only some or none of the arguments acquire the CenterDot head):

CenterDot[x, 3] /. CenterDot[a_, 3] :> {a, 3}
CenterDot[x, 3] /. CenterDot[a_, b_Integer] :> {a, b}
CenterDot[x, f[3]] /. CenterDot[a_, f[b_]] :> {a, b}
CenterDot[x, f[3]] /. CenterDot[a_, b_f] :> {a, First @ b}

{CenterDot[x], 3}

{x, 3}

{CenterDot[x], 3}

{x, 3}

Your rule is similar to the 3rd example above:

CenterDot[a, f[b, c]] /. CenterDot[a_, f[b_, c_]] :> {a, b, c}

{CenterDot[a], b, c}

Now, compare this to a version where CenterDot has both the Flat and OneIdentity attributes:

ClearAll[CenterDot]
SetAttributes[CenterDot, {Flat, OneIdentity}]

CenterDot[x, y] /. CenterDot[a_, b_] :> {a,b}

{x, y}

Note that the arguments have not acquired the CenterDot head. Now, your rule behaves the way you want:

CenterDot[a, f[b, c]] /. CenterDot[a_, f[b_, c_]] :> {a, b, c}

{a, b, c}

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  • $\begingroup$ Many thanks! This does work, but I do not quite understand where was the bug and how OneIdentity resolves the problem. $\endgroup$ – John Aug 23 '18 at 6:10
  • $\begingroup$ @John See update $\endgroup$ – Carl Woll Aug 23 '18 at 18:20
  • $\begingroup$ Dear Carl, many thanks! I am still confused, because what is above is a collection of examples, rather than an explanation for why the outputs are like these. It looks like with only 'Flat' attribute any replacement rule gives an output where CenterDot Head is randomly distributed :) $\endgroup$ – John Aug 24 '18 at 12:21

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