Find files older than FILE without including FILE

I have a plan to use `find` to give me a list of files that are older than some FILE and then use `xargs` (or `-exec`) to move the files to somewhere else. Moving stuff is not a problem; `| xargs mv {} ~/trash` works fine. Now, if I try to use `! -newer FILE`, then `FILE` is included in the list, which I do not want! The functionality of that command argument does indeed make sense logically, though, because 'not newer' could very well be interpreted as "same or older", like here: ``` $ find . ! -newer Selection_008.png -exec ls -l {} \; ``` includes the file from the compare argument: ``` -rw-r--r-- 1 and and 178058 2012-09-24 11:46 ./Selection_004.png -rw-r--r-- 1 and and 16260 2012-09-21 11:25 ./Selection_003.png -rw-r--r-- 1 and and 38329 2012-10-04 17:13 ./Selection_008.png -rw-r--r-- 1 and and 177615 2012-09-24 11:53 ./Selection_005.png ``` (`ls -l` is only to show dates for illustrative purposes) What I really need from `find` is an `-older` option, but none is listed in `find(1)`... Of course, one could just pipe the output through `grep -v`, use `sed`, etc., or utilise an environment variable to reuse the filename (fx. for `grep -v`) so I can enjoy the DRY-principle, like ``` $ COMP=Selection_008.png find . ! -newer $COMP | grep -v $COMP | xargs ... ``` but it just seems to be a lot of writing for a oneliner and that is not what I am looking for. Is there a shorter/simpler way than `find` or am I missing some option? I have checked the manpage, and searched Google and SO...