Suppose a list of homogeneous factors and
a list of heterogeneous factors constitute a tidy
factorization of .
If is either a convergent variable, or an old regular variable,
or a new regular variable whose deputy is not in the
then the procedure
a tidy heterogeneous factorization
Proof: Suppose , ..., are all the heterogeneous factors and , ..., are all the homogeneous factors. Also suppose , ..., , , ..., are all the factors that contain . Then
where equation (10) is due to
Proposition 3. Equation (9) is follows from Proposition
4. As a matter of fact,
if is a convergent variable, then it is the only convergent
variable in due to the first condition of
tidiness. The condition of Proposition 4 is satisfied because
does not appear in , ..., .
On the other hand, if is an old regular variable or a
new regular variable
whose deputy does not appear in
the list , ..., ,
contains no convergent variables due to the second
condition of tidiness. Again the condition of Proposition 4 is satisfied.
We have thus proved that
sum-out1 yields a
flexible heterogeneous factorization of .
Let be a convergent variable in the list , ..., .
Then cannot be the corresponding new regular variable e.
Hence the factor is not touched by
Consequently, if we can show that the new factor created
sum-out1 is either
a heterogeneous factor or a homogeneous factor that contain
no convergent variable, then the factorization
returned is tidy.
does not create a new homogeneous factor. Then no heterogeneous
factors in contain . If is a convergent variable, say ,
then is the only homogeneous factor that
contain . The new factor is , which does
contain any convergent variables. If is an old regular variable
or a new regular variable whose deputy is not in the list
, ..., ,
all the factors that contain do not contain any convergent variables.
Hence the new factor again does not contain any convergent variables.
The theorem is thus proved.