Correct state row entries in the following Sequential machine?

[elias001]

TL;DR

I would like to know if my calculations for the state row entries for $\{q_0,q_1,q_2,q_3,q_4,q_5\}$ are correct.

The two screenshots below is taken from the text Arrows, Structures, and functiors The categorical Imperative By: Arbib and manes, pp 93-94

page 1

page 2

In the above two screenshots, I would like to fill out the last state row entries: $q_0,q_1,q_2,q_3,q_4$

Let $t=\{0,1,2,3,4,5\}, Y=D=\{0,1,2,3,4,5,6,7,8,9\}, Q=\{0,1\}\times D$

$q_t=q(t)=(c,cd)=(c,d), c\in \{0,1\}, cd=10c+d, q_t: cd\mapsto d\in D$

$\beta:\{0,1\}\times D\to D:cd\mapsto d$

$\delta:\{0,1\}\times D\times D\times D\to \{0,1\}\times D:(cd,x_1,x_2)\mapsto c'd'=c+x_1+x_2$

where $x_1$ denotes $1$st input, and $x_2$ denotes $2$nd input. For the $\delta$ map, we can let $c'_{t+1}d_t'=c+x_{t1}+x_{t2},$ where $x_1=x_{t1}, x_2=x_{t2},$

Instead of $c'$, we also can let $c'=c'_{t+1}$ and $c_t=c\in \{0,1\}, c_{t+1}=1 \text{ if } c_t+x_{t1}+x_{t2}\geq 10, \text{ otherwise } c_{t+1}=0, \text{ also we let } d'=d_t.$

Then for:

$t=0, q_0=q(0)=(0,0,0),$ so $c_1=0, d_1=0, x_{01}=0, x_{02}=0, q_0=(0,0),$

$t=1, q_1=q(1)=(0,2,1)=(0,2+1), 0+2+4=3<10.$ So $c'_2=0,d'_1=3, x_{11}=2, x_{12}=1, q_1=(0,3),$

$t=2, q_2=q(2)=(0,7,4)=(0,7+4), 0+7+4=11>10.$ So $c'_3=1,d'_2=11-10=1, x_{21}=7, x_{22}=4,q_2=(0,1),$

$t=3, q_3=q(3)=(1,3,0)=(1,3+0), 1+3+0=4<10.$ So $c'_4=0,d'_2=4, x_{31}=3, x_{32}=0,q_3=(1,3),$

$t=4, q_4=q(4)=(0,0,0),$ so $c_5=0, d_4=0, x_{41}=0, x_{42}=0,q_4=(0,0),$

$t=5, q_5=q(5)=(0,0,0),$ so $c_6=0, d_5=0, x_{51}=0, x_{52}=0,q_5=(0,0).$

Thank you in advance