495 us have a look how the right-hand sides of the equations are |
495 us have a look how the right-hand sides of the equations are |
496 constructed. First have a look at the second equation: the |
496 constructed. First have a look at the second equation: the |
497 left-hand side is $q_1$ and the right-hand side $q_0\,a$. The |
497 left-hand side is $q_1$ and the right-hand side $q_0\,a$. The |
498 right-hand side is essentially all possible ways how to end up |
498 right-hand side is essentially all possible ways how to end up |
499 in $q_1$. There is only one incoming edge from $q_0$ consuming |
499 in $q_1$. There is only one incoming edge from $q_0$ consuming |
500 an $a$. Therefore we say: if we are in $q_0$ consuming an $a$ |
500 an $a$. Therefore the right hand side is |
501 then we end up in $q_1$. Therefore the right hand side is |
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502 state followed by character---in this case $q_0\,a$. Now lets |
501 state followed by character---in this case $q_0\,a$. Now lets |
503 have a look at the third equation: there are two incoming |
502 have a look at the third equation: there are two incoming |
504 edges. Therefore we have two terms, namely $q_1\,a$ and |
503 edges. Therefore we have two terms, namely $q_1\,a$ and |
505 $q_2\,a$. These terms are separated by $+$. The first states |
504 $q_2\,a$. These terms are separated by $+$. The first states |
506 that if in state $q_1$ consuming an $a$ will bring you to |
505 that if in state $q_1$ consuming an $a$ will bring you to |