1353 Q_0 & = & \ONE + Q_0\,(b + a\,b) + Q_2\,b\\ |
1353 Q_0 & = & \ONE + Q_0\,(b + a\,b) + Q_2\,b\\ |
1354 Q_2 & = & Q_0\,a\,a + Q_2\,a |
1354 Q_2 & = & Q_0\,a\,a + Q_2\,a |
1355 \end{eqnarray} |
1355 \end{eqnarray} |
1356 |
1356 |
1357 \noindent Unfortunately we cannot make any more progress with |
1357 \noindent Unfortunately we cannot make any more progress with |
1358 substituting equations, because both (6) and (7) contain the |
1358 substituting equations, because both (8) and (9) contain the |
1359 variable on the left-hand side also on the right-hand side. |
1359 variable on the left-hand side also on the right-hand side. |
1360 Here we need to now use a law that is different from the usual |
1360 Here we need to now use a law that is different from the usual |
1361 laws about linear equations. It is called \emph{Arden's rule}. |
1361 laws about linear equations. It is called \emph{Arden's rule}. |
1362 It states that if an equation is of the form $q = q\,r + s$ |
1362 It states that if an equation is of the form $q = q\,r + s$ |
1363 then it can be transformed to $q = s\, r^*$. Since we can |
1363 then it can be transformed to $q = s\, r^*$. Since we can |
1364 assume $+$ is symmetric, Equation (7) is of that form: $s$ is |
1364 assume $+$ is symmetric, Equation (9) is of that form: $s$ is |
1365 $Q_0\,a\,a$ and $r$ is $a$. That means we can transform |
1365 $Q_0\,a\,a$ and $r$ is $a$. That means we can transform |
1366 (7) to obtain the two new equations |
1366 (9) to obtain the two new equations |
1367 |
1367 |
1368 \begin{eqnarray} |
1368 \begin{eqnarray} |
1369 Q_0 & = & \ONE + Q_0\,(b + a\,b) + Q_2\,b\\ |
1369 Q_0 & = & \ONE + Q_0\,(b + a\,b) + Q_2\,b\\ |
1370 Q_2 & = & Q_0\,a\,a\,(a^*) |
1370 Q_2 & = & Q_0\,a\,a\,(a^*) |
1371 \end{eqnarray} |
1371 \end{eqnarray} |