Theorem
3
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
list 
,
then the procedure sum-out1
returns
a tidy heterogeneous factorization
of 
.
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 
, ..., 
,
then
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
sum-out1
.
Consequently, if we can show that the new factor created
by sum-out1
is either
a heterogeneous factor or a homogeneous factor that contain
no convergent variable, then the factorization
returned is tidy.
Suppose
sum-out1
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. 
