I am re-writing the question using a very simple example so not to confuse matters with the another question being asked where I used the example from there.

In general, how to add logic to switch between one TransformationFunctions or another inside Simplify or any function that uses TransformationFunctions?

For example,

ClearAll[n, tf, e, cf];
cf[e_] := LeafCount@e
tfForIntegerOnly[e_] := If[MatchQ[e, 2 n], n, e]
tfForRealOnly[e_] := If[MatchQ[e, 2 n], n^2, e]

Assuming[Element[n, Integers], 
 Simplify[2 n, TransformationFunctions -> {Automatic, tf}, ComplexityFunction -> cf]]

How to make it use tfForIntegerOnly when Element[n, Integers] and use tfForRealOnly when Element[n, Reals]

Where to add this logic? How to pass the assumptions around or check for it and what it contains? Since Mathematica does not have types (in traditional sense) associated or attached with the symbols themselves (other than looking at Head, which in this case provides no information), one needs a general way to handle this.

Original question below

I'd like to simplify an expression under one set of assumptions using one TransformationFunctions function and use another TransformationFunctions function (or use Automatic) for different set of assumptions.

The problem is that the TransformationFunctions itself has no access to these Assumptions used before in calling Simplify and hence it is hard to find a way to add logic to detect which assumptions is used at that level.

To explain, I want to apply TransformationFunctions to transform Gamma[1/2 + n] to Sqrt[Pi] (Factorial2[2 *(n - 1/2) - 1])/2^(n - 1/2) but only when Element[n, Integers] && n > 1.

I am not able to find a way to pass these assumptions from the Assuming call to the TransformationFunctions or a way to make 2 different TransformationFunctions and use vs. the other inside Simplify itself.

Here an example

ClearAll[n, tf, e, cf];
cf[e_] := Count[e, Gamma, Infinity, Heads -> True] 1000 + LeafCount@e 
tf[e_] := If[MatchQ[e, Gamma[1/2 + n]],Sqrt[Pi] (Factorial2[2 *(n - 1/2) - 1])/2^(n - 1/2), e]

Assuming[Element[n, Integers] && n > 1,Simplify[Gamma[1/2 + n] + Gamma[1/3 + n], 
TransformationFunctions -> {Automatic, tf},ComplexityFunction -> cf]]

Mathematica graphics

But now if I assume n is Real, I do not want to modify the code above, but I want the TransformationFunctions to automatically to detect this and in this case not apply this transformation rule or add some logic inside Simplify to use one TransformationFunctions vs. the other based on Assumptions

ClearAll[n, tf, e, cf];
cf[e_] := Count[e, Gamma, Infinity, Heads -> True] 1000 + LeafCount@e 
tf[e_] := If[MatchQ[e, Gamma[1/2 + n]], Sqrt[Pi] (Factorial2[2 *(n - 1/2) - 1])/2^(n - 1/2), e]

Assuming[Element[n, Reals], Simplify[Gamma[1/2 + n] + Gamma[1/3 + n], 
  TransformationFunctions -> {Automatic, tf}, 
  ComplexityFunction -> cf]]

Mathematica graphics

which is wrong since now n is not an integer now.

I'd like to be able to write

Assuming[Element[n, Reals], Simplify[Gamma[1/2 + n] + Gamma[1/3 + n],
TransformationFunctions ->Cases[Element[n, Reals], Automatic, Element[n, Integers] && n > 1,tf],
  ComplexityFunction -> cf]

But the above is not valid syntax.

  • $\begingroup$ see my answer on the original question. $\endgroup$ Aug 12, 2013 at 21:33
  • $\begingroup$ I got in the same problem when dealing with the original question too. At the time, I noticed that Assuming[n \[Element] Integers, n \[Element] Integers] returns n \[Element] Integers but Assuming[n \[Element] Integers, Simplify[n \[Element] Integers]] returns True. Reading the manual, it says that Assuming is used only for certain functions (Simplify, Integrate, …) $\endgroup$
    – Hector
    Aug 12, 2013 at 22:01
  • $\begingroup$ And again, my answer can be used for that. Have you actually studied it? All you need is a different condition in the FullSimplify part. The condition can also be placed at the at of the rule if that's more convenient. $\endgroup$ Aug 12, 2013 at 22:23
  • $\begingroup$ It is very easy to generalize my answer to any rule you like. Just use the same transformation function name name for all your rule based transformations and use function overloading by varying the condition /; part. Also, you haven't read my comment about the use of the global symbol n. $\endgroup$ Aug 12, 2013 at 22:59
  • $\begingroup$ Btw, could you use the @ convention when replying? I'll be away for the rest of the night. Bye. $\endgroup$ Aug 12, 2013 at 23:04

1 Answer 1


I had a bit of trouble following the use of explicit (hard-coded) Symbols (n) in your transformation functions. I assumed these were used only for the simple example and replaced them with patters in my code below.

  • One issue that may be causing you trouble is that Integers is a subset of Reals so you need to test for the subset rather than the superset.

  • Since Simplify is responsive to $Assumptions and by extension Assuming we can use it to check our Condition.

I believe you want something like this:

ClearAll[n, tf, e, cf];
cf[e_] := Count[e, Gamma, Infinity, Heads -> True] 1000 + LeafCount@e

tf[Gamma[1/2 + x_]] /; Simplify[x ∈ Integers] :=
  Sqrt[Pi] (Factorial2[2*(x - 1/2) - 1])/2^(x - 1/2)


  Assuming[Element[n, domain] && n > 1, 
    Simplify[Gamma[1/2 + n] + Gamma[1/3 + n],
      TransformationFunctions -> {Automatic, tf}, 
      ComplexityFunction -> cf]],
  {domain, {Integers, Reals}}
] // Column
2^(1/2 - n) Sqrt[π] (2 (-1 + n))!! + Gamma[1/3 + n]

Gamma[1/3 + n] + Gamma[1/2 + n]
  • $\begingroup$ Isn't this equivalent to the method I linked to in my comment above? $\endgroup$ Aug 14, 2013 at 15:32
  • $\begingroup$ @Sjoerd I suppose it is, but I didn't see it until now. Whatever, I hope this helps the OP solve his problem. $\endgroup$
    – Mr.Wizard
    Aug 14, 2013 at 15:54

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