order of operations: erase old files before building new ones - makefile

Question

Although I though I was pretty good with makefiles, I have a problem I just can't solve about order of commands in a makefile.
Lets detail.
Say I have some program that produces some files of type "a". I want to manually take these and process them through some tool that takes as input one "a" file and produces a file of type "b".
So my makefile holds the following pattern rule:
%.b:%.a
$(MYTOOL) $< >$#
I gather all the files to build in a variable B_FILES, build from the list of "a" files:
A_FILES = $(wildcard *.a)
B_FILES = $(patsubst %.a,%.b,$(A_FILES))
And this is called through a general call target:
build_b_files: $(B_FILES)
#echo "done!"
Works perfectly.
Now the problem: each run of the main program can generate a varying number of "A" files, with some common naming scheme (say file_0.a, file_1.a, file_2.a, ...).
But if a run produces 20 files, and the following only 10, then I still have the remaining 10 files from latter run present.
So what I want to do is make my build_b_files target automatically erase all the old "b" file before building the new ones.
I tried to use the order-only prerequisite, but the following fails: it just rebuilds every time all the files, then erases them (!).
erase_all_B:
#rm *.b
build_b_files: $(B_FILES) | erase_all_B
#echo "- Done target $#"
I am probably misunderstanding something about this concept of "order-only".
How can I achieve this goal ?
Edit: to clarify question, see the following (simulated) session demonstrating what I am looking for (ls only showing relevant files here):
$ myapp some-arguments... (will produce 4 'a' files)
$ ls -l (stripped)
june 8 16:00 file_0.a
june 8 16:00 file_1.a
june 8 16:00 file_2.a
june 8 16:00 file_3.a
$ make build_b_files
$ ls -l
june 8 16:00 file_0.a
june 8 16:00 file_1.a
june 8 16:00 file_2.a
june 8 16:00 file_3.a
june 8 16:00 file_0.b
june 8 16:00 file_1.b
june 8 16:00 file_2.b
june 8 16:00 file_3.b
.... (some time) ....
$ myapp some-other-arguments... (will produce only 3 'a' files)
$ ls -l
june 8 17:00 file_0.a
june 8 17:00 file_1.a
june 8 17:00 file_2.a
june 8 16:00 file_3.a
june 8 16:00 file_0.b
june 8 16:00 file_1.b
june 8 16:00 file_2.b
june 8 16:00 file_3.b
$ make build_b_files
$ ls -l
june 8 17:00 file_0.a
june 8 17:00 file_1.a
june 8 17:00 file_2.a
june 8 17:00 file_0.b
june 8 17:00 file_1.b
june 8 17:00 file_2.b
(For those who wonder, this is about graphviz, that generates images from dot files...)

Answer 1

You need to erase all b files before remaking any b files, so what you're saying is that erase_all_B should be a prerequisite of the $(BFILES) themselves, not build_b_files:
.PHONY: erase_all_B
erase_all_B: ; $(RM) *.b
$(B_FILES): erase_all_B
Test with the following
all: bar baz
bar baz: foo
ba%:
#echo bar baz
foo:
#echo foo

Answer 2

If the tool that produces .a files removes the old ones, you can build a "list of .b files to remove" easily with something like:
TO_REMOVE := $(filter-out $(B_FILES),$(wildcard *.b))
And if the tool that produces .a files does not remove older ones, why removing matching .b files? They will be regenerated, as .a files are still here. And if .a files are modified, corresponding .b files will be regenerated by make (as being out of date).