Behavior of the function 'has'

12 Mar 2002


      I do not understand the behavior of the function 'has' in some case.
I show you the problem with ginsh:
...
has(2^x*3^x, $0^x);
1
has(2^x*3^x*4^x, $0^x);
1
has(2^x*3^x, $0^x*$1^x);
1
has(2^x*3^x*4^x, $0^x*$1^x);
0
has(2^x*3^x*4^x, $0^x*$1^x*$2^x);
1
I would expect that also

has(2^x*3^x*4^x, $0^x*$1^x);

answers 'true' because I see $0^x*$1^x as a subexpression of

2^x*3^x*4^x.


Same thing without to use wildcard.
...
has(2^x*3^x*4^x, 2^x);
1
has(2^x*3^x*4^x, 2^x*3^x);
0
has(2^x*3^x*4^x, 2^x*3^x*4^x);
1
Regards

     Tatiana Zolo