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