13 Mar
2002
13 Mar
'02
3:49 p.m.
Hi! On Tue, Mar 12, 2002 at 12:59:07PM +0100, Tatiana Zolo wrote:
has(2^x*3^x*4^x, 2^x); 1 has(2^x*3^x*4^x, 2^x*3^x); 0
has(e, x) returns true if and only if a subexpression of e (as defined by op()) matches (as defined by match()) x (or e matches x entirely). "2^x*3^x" is not a subexpression of "2^x*3^x*4^x" by this definition. A workaround is to use
has(2^x*3^x*4^x, 2^x*3^x*$0); 1
Bye, Christian -- / Coding on PowerPC and proud of it \/ http://www.uni-mainz.de/~bauec002/