Q$2.$ It is said that the Boolean functions of the output variables of the circuit shown in Fig. Q$2$ are

given as

$-T_1\frac{?}{b}=ar(B)C,*T_2\frac{?}{b}=ar(A)B,*T_3=A+T_1=A\frac{+}{b}ar(B)C$

$-T_4=T_{2}o+D=(\frac{?}{\text{b}\text{a}\text{r}}(A)B)o+D\frac{?}{b}=ar(A)\frac{B}{b}ar(D)+AD\frac{+}{b}ar(B)D$

$-F_1=T_3+T_4=A\frac{+}{b}ar(B)C\frac{+}{b}ar(A)\frac{B}{b}ar(D)+AD\frac{+}{b}ar(B)D$

$=A\frac{+}{b}ar(B)C+\frac{B}{b}ar(D)\frac{+}{b}ar(B)D$

$-F_2=T_2+D\frac{?}{b}=ar(A)B+D$

Prove that the following equation is true:

$F_1-=A\frac{+}{b}ar(B)C\frac{+}{b}ar(A)\frac{B}{b}ar(D)+AD\frac{+}{b}ar(B)D=A\frac{+}{b}ar(B)C+\frac{B}{b}ar(D)\frac{+}{b}ar(B)D$