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Re: FullForm puzzle.

  • To: mathgroup at
  • Subject: [mg106302] Re: FullForm puzzle.
  • From: Norbert Marxer <marxer at>
  • Date: Fri, 8 Jan 2010 04:12:41 -0500 (EST)
  • References: <hi42dp$n3h$>

On Jan 7, 8:27 am, Jack L Goldberg 1 <jackg... at> wrote:
> Fellow mathematistas:
> I sent this post to the two responders but thought it might be of interest
>   I use a transformations in some of my code that takes a <= x <= b
> and converts it into a function I have defined called "characteristic  
> function" - the name is obviously irrelevant.  But to make this  
> transformation I need to know what the FullForm is. If it (the  
> FullForm) varies from one command to another, how can I count on my  
> transformation working.  For example, suppose a line in my Module  
> looks like this:
> (1)        "something"/.LessEqual[2,x,3] -> characteristic[x,2,3]
> But if "something" contains the FullForm "Inequality[ ... ]" the  
> transformation explicit in (1) will fail.
> I ran into this and found a simple work-around.  However, it took me a 
> couple of hours to pinpoint the problem since it was imbedded in a  
> relatively long code.
> Jack


I would use LogicalExpand with your expressions. Then your three
expressions will all have the same FullForm.

(*This gives False*)
e1 = LessEqual[2, x, 3]; e2 =  Inequality[2, LessEqual, x, LessEqual,
3]; e3 = 2 <= x && x <= 3;
SameQ[e1, e2, e3]

(*This gives True*)
e1L = LogicalExpand[e1]; e2L = LogicalExpand[e2]; e3L =  LogicalExpand
SameQ[e1L, e2L, e3L]
SameQ[e1L // FullForm, e2L // FullForm, e3L // FullForm]

I hope this helps.

Best Regards
Norbert Marxer

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