Indeed what we have proved is that if (HGRAIL) has a solution with invertible and if b and c are given by formulas (6.11) and (6.13), then
Now we turn to the converse. The straightforward converse of the above italicized statement would be: If items (1), (2) and (3) above hold, then (HGRAIL) has a solution with invertible and b and c are given by formulas (6.11) and (6.13). There is no reason to believe (and it is not the case) that b and c must be given by the formulas (6.11) and (6.13). These two formulas came about in § and were motivated by ``excess freedom'' in the problem. The converse which we will attempt to prove is:
Proposed Converse 6.14 If items (1), (2) and (3) above hold, then (HGRAIL) has a solution with invertible.
To obtain this proposed converse, we need a complete spreadsheet corresponding to the last stages of our analysis. The complete spreadsheet is:
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YOUR SESSION HAS DIGESTED  

THE FOLLOWING RELATIONS  
a E A E + E C B E + E C B E + E E B B E  E C C E E  E E B C E E  E C B E E E  E E B B E E E
b E C  E E B
c B E + C E E + B E E E
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a E A E + E B C E + E B C E + E B B E E  E E C C E  E E E B C E  E E C B E E  E E E B B E E
b C E  B E E
c E B + E E C + E E E B
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Y B C Y A + A Y + C C  C C  C B Y
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USER CREATIONS APPEAR BELOW 
 
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SOME RELATIONS WHICH APPEAR BELOW 
 
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In the spreadsheet, we use conventional X, notation rather than ``discovered'' notation so that our arguments will be familiar to experts in the field of control theory.
Now we use the above spreadsheet to verify the proposed converse. To do this, we assume that matrices A, , , , , X and Y exist, that X and Y is invertible, that X and Y are selfadjoint, that is invertible and that the DGKF X and equations hold. That is, the two following polynomial equations hold.
We wish to assign values for the matrices , , , , a, b and c such that each of the equations on the above spreadsheet hold. If we can do this, then each of the equations from the starting polynomial equations from § will hold and the proposed converse will follow.
With the assignments of , , , , a, b and c as above, it is easy to verify by inspection that every polynomial equation on the spreadsheet above holds.
We have proven the proposed converse and, therefore, have proven the following approximation to the classical [DGKF] theorem.
Theorem 6.15 If (HGRAIL) has a solution with invertible and b and c are given by the formulas (6.11) and (6.13), then the DGKF X and Y equations have solutions X and Y which are symmetric matrices with X, Y and Y  X invertible. The DGKF X and Y equations have solutions X and Y which are symmetric matrices with X, Y and Y  X invertible, then (HGRAIL) has a solution with invertible .
Note that we obtained this result with an equation in the one unknown X and an equation with the one unknown . From the strategy point of view, the first spreadsheet featured an equation in the single unknown b (and its transpose) and an equation in the single unknown c (and its transpose) and so is the most complicated. For example, the (6.10) decomposes as
where and is a symmetric polynomial which does not involve b. This forces us to say that the proof of the necessary side of theorem 6.15 was done with a 2strategy.
A more aggressive way of selecting knowns and unknowns allows us to obtain this same result with a symmetrized 1strategy. In particular, one would set a, b and c to be the only unknowns to obtain a first spreadsheet. The first spreadsheet contains key equations like (6.16) which is a symmetric 1decomposition because does not contain a, b or c. Once we have solved for a, b and c, we turn to the next spreadsheet by declaring the variables involving (e.g., , , ...) to be unknown. At this point, the computer run is the same as Steps 2, 3 and 4 above.