I am running symbolic calculations with Mathematica. Simplification of algebraic expressions is an important part of them as very often final results can be simplified considerably. Typically my expressions involve many variables that can be categorised into few groups. I use to denote a group by a letter, variables within the group are distinguished by an index, e.g. $A_1$, $B_2$, etc. At some point I found out (and tried to discuss it here Simplifying using assumptions) that FullSimplify command best works with variables without any subscripts. In the example below the result of the simplification is cardinally different depending whether I am using plain letters


or letters with subscripts

$\frac{B_1 \left(a_1+x\right) \left(A_2 \left(a_2-x\right) \left(-a_1+a_2+2 x\right)-\left(a_1-a_2\right) B_2 \left(a_2+x\right)\right)-A_1 \left(a_1-x\right) \left(\left(a_1-a_2\right) A_2 \left(a_2-x\right)+B_2 \left(a_1-a_2+2 x\right) \left(a_2+x\right)\right)}{\left(x-a_1\right) \left(a_1+x\right) \left(a_2-x\right) \left(a_2+x\right)}$.

Clearly, I prefer the first form. Therefore I am using rules to replace indexed variables with plain letters, perform a simplification, and using the rules again to replace plain letters by the original indexed variables. However, it is not very convenient method, also in view of the fact that the number of letters is finite...

Therefore I have a question, is there a way to teach FullSimplify to work as efficiently with indexed variables as it does with the normal ones?

Below are two equivalent expressions. The first one contains only letters

f2 = (-p (a - x) ((a - b) P (b - x) + Q (b + x) (a - b + 2 x)) + q (a + x) (-(a - b) Q (b + x) + P (b - x) (-a + b + 2 x)))/((b - x) (-a + x) (a + x) (b + x));
rule2 = {(p + q) -> 1, (P + Q) -> 1};
FullSimplify[f2] /. rule2

It leads to the first equation. The second one

f1 = ((x + Subscript[a,1]) Subscript[B,1]((-x + Subscript[a, 2])(2 x - Subscript[a,1] + Subscript[a,2]) Subscript[A,2]-(Subscript[a,1]-Subscript[a, 2]) (x + Subscript[a,2]) Subscript[B,2]) - (-x + Subscript[a,1]) Subscript[A,1] ((Subscript[a, 1] - Subscript[a,2]) (-x + Subscript[a,2]) Subscript[A,2] + (2 x + Subscript[a, 1] - Subscript[a, 2]) (x + Subscript[a,2]) Subscript[B,2]))/((x - Subscript[a,1]) (x + Subscript[a,1]) (-x + Subscript[a, 2]) (x + Subscript[a, 2]));
rule1 = {Subscript[A,1] + Subscript[B,1] -> 1, Subscript[A,2] + Subscript[B,2] -> 1};
FullSimplify[f1] /. rule1

Is equivalent to the first one, however, it is returned unsimplified (at least with my version of Mathematica: 9.0.1 for Mac OS X). Two expressions are equivalent as the following command shows

FullSimplify[f1 - f2 /. {Subscript[a,1] -> a, Subscript[a,2] -> b, Subscript[A,1] -> p, Subscript[A,2] -> P, Subscript[B,1] -> q,Subscript[B,2] -> Q}]

Returns 0.

Thank you in advance.

  • 3
    $\begingroup$ Why can't you use a1, a2, a3... b1, b2, ..., z1, z2, ... aa1,... ab1...? You have an infinite pool of symbols! $\endgroup$
    – rm -rf
    Feb 3, 2014 at 16:34
  • $\begingroup$ How to define an order relation on your set? $\endgroup$
    – yarchik
    Feb 4, 2014 at 9:23
  • $\begingroup$ There seems to be a sort of side-question (which @rm-rf may want to weigh in on) where you want to use subscripts in "looping constructs" and to make things "visually appealing". For the latter, you can convert at the end to whatever presentation you want. For the former, it might be better to use things like a[1], a[2],... but maybe this is worthy of a search and possibly a separate question about the best way to handle indices for the looping constructs you have in mind. $\endgroup$
    – Mark S.
    Feb 5, 2014 at 4:15
  • $\begingroup$ @ Mark S. Well, have you tried before suggesting? "Things like a[1], a[2], ... " do not work either ... A suggestion of rm -rf does work, but, as I said, additional work must be invested in implementing looping with these variables. I do not say that suggestion rm -rf is not a plausible solution: It is just preaching instead of answering a precisely formulated question. $\endgroup$
    – yarchik
    Feb 5, 2014 at 10:03

3 Answers 3


I agree with point 5 of Verbeia's answer to the canonical Q&A, i.e. you should avoid subscripted variables. One solution here might be to automatically convert "subscript Symbols" to regular symbols and to convert them back after having using FullSimplify. The function below does this.

fullSimplify[expr_] :=
  {uqString, counter, definedQ, toMapper, backRules, symb, simpExpr, 

  backRules = {};
  definedQ[x__] := False;
  uqString = ToString[Unique[bla]];
  counter = 0;

   Subscript[x__] := 
      definedQ[x] = True;
      symb = Symbol[uqString <> ToString[counter]];
       symb :> Subscript[x]
      toMapper[x] = symb
   simpExpr =
  Remove[toMapper, definedQ];
  simpExpr /. backRules



fullSimplify[f1] /. rule1


enter image description here


It is better to use a linkedlist than to use a regular list with AppendTo

The variables from Module will leak unless they are removed.

  • $\begingroup$ This answers my question although it does not improve my understanding of inner mechanisms of Mathematica. I am not making any assignments to the subscripted variables. Should not they have all the properties of complex numbers? It is hard for me to accept your recommendation of completely avoiding subscripted variables: they are very useful in looping constructs. $\endgroup$
    – yarchik
    Feb 4, 2014 at 9:20
  • $\begingroup$ @yarchik I am not sure why the same simplification is not made. You should know the measure for simplicity used by FullSimplify is LeafCount. The LeafCount of a subscripted variable is different than that of a normal Symbol, so this may cause different behaviour. I do not really understand your comment about looking constructs, could you elaborate? $\endgroup$ Feb 4, 2014 at 11:13
  • $\begingroup$ @yarchik If you think this is bad, have you seen Variable naming changes everything? Even a choice of a different letter changes behavior in other contexts. $\endgroup$
    – Mark S.
    Feb 5, 2014 at 4:10
  • $\begingroup$ @Mark S. I haven't seen this discussion before, but now I am amazed... $\endgroup$
    – yarchik
    Feb 5, 2014 at 9:11
  • $\begingroup$ @Jacob Akkerboom I have tried to follow your suggestion and used a different complexity function that disregards Subscripts: comp[e_] := -2 Count[e, _Subscript, {0,Infinity}] + LeafCount[e] However, it does not help. $\endgroup$
    – yarchik
    Feb 5, 2014 at 10:33

If you follow rm -rf advice, everything goes smoothly:

expr = 
  (-A1 (a1 - x) ((a1 - a2) A2 (a2 - x) + B2 (a2 + x) (a1 - a2 + 2 x)) + 
    B1 (a1 + x) (-(a1 - a2) B2 (a2 + x) + A2 (a2 - x) (-a1 + a2 + 2 x))) /
  ((a2 - x) (-a1 + x) (a1 + x) (a2 + x));
rule = {A1 + B1 -> 1, A2 + B2 -> 1};
FullSimplify[expr] /. rule


  • $\begingroup$ Well, I cannot accept this answer simply because it does not answer the question: there are many cases when subscripts are useful and to work with your variables is pretty inconvenient. Second, visually it is less appealing. $\endgroup$
    – yarchik
    Feb 4, 2014 at 9:15
  • 1
    $\begingroup$ @yarchik As with everything, there's a trade off and you need to make a decision — do you want to obtain the solution (without additional effort) or do you want something that is simply visually appealing? You might find a solution to this problem, but using subscripts heavily is almost sure to end in tears. $\endgroup$
    – rm -rf
    Feb 4, 2014 at 15:07

Symbolizing subscripts before running your code also works.

<< Notation`
SubscriptBox["_", "_"]]]

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.