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.