I have a list of functions, each containing none or more of the functions Plus, Times, Subtract and Divide. They could be in any arrangement and with any numbers. This is one such example.


There is an error here, The above should read: Plus[3,Times[7

The numbers can be any positive integer. I need to calculate just the Divide and Subtract functions to find out if they evaluate to 1 or 0 respectively as I consider functions in the list that have this property trivial and I want to filter them out.

I am somewhat a beginner with Wolfram/Mathematica code and I have been attempting to extract the Divide and Subtract parts as a list and then calculating just them parts but so-far my attempts at solving this have been totally unfruitful, partly because I can't know beforehand where the Divide or Subtract functions are in the function or how deep they will be.

Is there a way to accomplish this?

Edit: Just to help clarify what I am trying to achieve. I have a list of functions such as

{HoldForm[Plus[3,Times[7,Subtract[9,Plus[4,5]],Divide[29,Subtract[4,3]]]]], HoldForm[Plus[7,Times[2,Divide[16,Subtract[9,Plus[4,1]]]]]], HoldForm[Plus[3,Times[7,9],29]]}

I need to remove the first element from the list because it is a function that divides by 1 (and subtracts 0), making is only trivially different from the 3rd element of the list. After removing the first function I want to display the output like this

{$7+2 \frac{16}{9-(4+1)}$,$3+7\times 9+29$}

  • $\begingroup$ Welcome to Mathematica.SE! Can you please put any code snippets in code blocks to improve the readability of the question? $\endgroup$
    – Szabolcs
    Commented Sep 13, 2018 at 13:05
  • 1
    $\begingroup$ What would be a result of Subtract[9,Plus[4,5]] if Plus is not included in your list? $\endgroup$
    – Kuba
    Commented Sep 13, 2018 at 13:10
  • $\begingroup$ In my example, the whole thing should be removed from my list of functions. This is because Subtract[9,Plus[4,5]] will evaluate to 0 and so it means the example is trivial as there will be another function in the list that is the same except it won't Subtract[0] $\endgroup$
    – RedPython
    Commented Sep 13, 2018 at 13:19
  • $\begingroup$ In that case why don't you just evaluate the whole thing and say that the result is 3? $\endgroup$
    – Szabolcs
    Commented Sep 13, 2018 at 13:21
  • $\begingroup$ Thanks for the advice Szabolcs. Advice appreciated. I have looked up how to do it and will ensure I follow your advice going forward. $\endgroup$
    – RedPython
    Commented Sep 13, 2018 at 13:21

2 Answers 2


Update: Selecting from a list expressions those that do not contain subexpressions that divide by 1 or subtract 0:

list = {HoldForm[Plus[3, Times[7, Subtract[9, Plus[4, 5]], Divide[29, Subtract[4, 3]]]]],
   HoldForm[Plus[7, Times[2, Divide[16, Subtract[9, Plus[4, 1]]]]]],
   HoldForm[Plus[3, Times[7, 9], 29]]};

condition = FreeQ[#, (s_Subtract /; s == 0) | (d_Divide /; d == 1),  ∞] &;
Select[list, condition] 

{7 + 2 16/(9 - (4 + 1)), 3 + 7 9 + 29}

enter image description here

Original answer:

exp = Inactivate[Plus[3, Times[7, Subtract[9, Plus[4, 5]], Divide[29, Subtract[4, 3]]]]];
Activate[exp, Divide | Subtract]

3 + 7*(9 - (4 + 5))*29

An alternative way to evaluate completely subexpressions with head Divide or Subtract using RuleCondition (from WReach's answer in the q/a linked in Sjoerd's answer):

HoldForm[Plus[3, Times[7, Subtract[9, Plus[4, 5]], Divide[29, Subtract[4, 3]]]]] /. 
    e : _Subtract | _Divide :> RuleCondition[e]

3 + 7 0 29


The answer of kglr works if you only want to do the evaluations associated with Divide and Subtract. If instead you want every subexpression with these head to evaluate completely, use the following trick:

   3 + Times[7, Subtract[9, Plus[4, 5]], 
     Divide[29, Subtract[4, 3]]]]
] /. {
   expr : (_Divide | _Subtract) :> With[{x = expr}, x /; True]

See also the following answer:

Replacement inside held expression


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