SOLUTION OF THE TASK

Sequential circuit gives the following relations, that help in finding the current state of the machine

new $B=$ old $A$

new $A=(input)$ NAND $(old$ A AND old $B)$

output $=(input)$ AND $(oldB)$

The transition at the state $q_0$ while reading the letter $0,$ can be determined, along with the corresponding output character as under

new $B=$oldA$=0$

new $A=(input)$ NAND $(old$ A AND old $B)$

$=0$ NAND $(0$ AND $0)=0$ NAND $0=1$

output $=(input)$ AND $(oldB)=0$ AND $0=0$

SOLUTION CONTINUED $...$

Thus after reading $0$ at $q_0$ new $B$ is $0$ and new $A$ is I i$.$e$.$ machine will be at state $(1,0)-=q_2$ and during this process it's output character will be $0.$

The remaining action of this sequential circuit can be determined as shown by the following suggested transition table of the corresponding Mealy machine

$\backslash$table$[[$Old state,Inputting $0,$Inputting $1],[,$State,Output,State,Output$],[q_0-=(0,0),(1,0)-=q_2,0,(1,0)-=q_2,0],[q_1-=(0,1),(1,0)-=q_2,0,(1,0)-=q_2,1],[q_2-=(1,0),(1,1)-=q_3,0,(1,1)-=q_3,0],[q_3-=(1,1),(1,1)-=q_3,0,(0,1)-=q_1,1]]$

The corresponding transition diagram may be as follows