Re: precedence for ReplaceAll?

*To*: mathgroup at smc.vnet.net*Subject*: [mg110559] Re: precedence for ReplaceAll?*From*: Szabolcs Horvát <szhorvat at gmail.com>*Date*: Fri, 25 Jun 2010 07:28:31 -0400 (EDT)*References*: <hvv4ti$fc$1@smc.vnet.net>

[note: message sent to comp.soft-sys.math.mathematica as well] On 2010.06.24. 10:27, Matthias Fripp wrote: > I am having trouble using ReplaceAll to replace symbols that already > have a delayed assignment. > > e.g., this input: > > In[287]:= > a := A c > b := B c > a /. {a -> x, b -> y} > b /. {a -> x, b -> y} > a + b /. {a -> x, b -> y} > a * b /. {a -> x, b -> y} > > gives this output: > > Out[289]= x > > Out[290]= y > > Out[291]= x + y > > Out[292]= A B c^2 > > All of this works as expected except for the final term. I would have > expected to get the result "x y". Is there any way to force > Mathematica to produce that result? > > If on the other hand the original assignment is a := A + c and b := B > + c, I get an unexpected output for the sum, but the expected output > (x y) for the product. If I insert d instead of one of the c's, I get > various other (unpredictable) kinds of result. > > My first guess is that Mathematica is doing a sort of "double > ReplaceAll", where it first tries the pattern given in the delayed > assignment, and any symbols matched by that are not tested against the > explicit ReplaceAll. But that doesn't explain why the sum works and > not the product. Am I thinking about this the wrong way? > I get the impression from your message that you believed that when evaluating a /. {a -> x, b -> y}, 'a' gets replaced by 'x'. This is not the case. First the 'a' on the lhs of /. is evaluated to A c, then the 'a' on the rhs is evaluated to A c (and 'b' is evaluated too, of course). At this point the evaluator has the expression A c /. { A c -> x, B c -> y }, and replaces A c by x. This is all easy to see by using the Trace function: In[41]:= Trace[a/.a->x] Out[41]= {{a,A c},{{a,A c},A c->x,A c->x},A c/.A c->x,x} What happens with the last expression is this: In[42]:= Trace[a b/.{a->x,b->y}] Out[42]= {{{a,A c},{b,B c},(A c) (B c),A c B c,A B c c,A B c^2},{{{a,A c},A c->x,A c->x},{{b,B c},B c->y,B c->y},{A c->x,B c->y}},A B c^2/.{A c->x,B c->y},A B c^2} I.e. the ReplaceAll expression evaluates to A B c^2 /. {A c->x,B c->y}, and no replacement can be done, as A c or B c do not appear *formally* in the expression A B c^2. What you are looking for is achieved by Unevaluated[a b] /. {HoldPattern[a] -> x, HoldPattern[b] -> y} Hope this helps, Szabolcs