Using Boolean algebra techniques, simplify the following expressions as much

as possible:

a$.A(A+B)$

b$.A(\frac{?}{\text{b}\text{a}\text{r}}(A)+AB)$

c$.BC\frac{+}{b}ar(B)C$

d$.A(A\frac{+}{b}ar(A)B)$

e$.\frac{A}{b}ar(B)C\frac{+}{b}ar(A)BC\frac{+}{b}ar(AB)C$a$.(A\frac{+}{b}ar(B))(A+C)$

b$.\frac{?}{\text{b}\text{a}\text{r}}(A)B\frac{+}{b}ar(A)\frac{B}{b}ar(C)\frac{+}{b}ar(A)BCD\frac{+}{b}ar(A)\frac{B}{b}ar(CD)E$

c$.AB\frac{+}{b}ar(AB)C+A$

d$.(A\frac{+}{b}ar(A))(AB+A\frac{B}{b}ar(C))$

e$.AB+(\frac{?}{\text{b}\text{a}\text{r}}(A)\frac{+}{b}ar(B))C+AB$

Using Boolean algebra, simplify each expression:

a$.BD+B(D+E)\frac{+}{b}ar(D)D+F$

b$.\frac{?}{\text{b}\text{a}\text{r}}(AB)C\frac{+}{b}ar((A+B+(\frac{?}{\text{b}\text{a}\text{r}}(C))))\frac{+}{b}ar\frac{A}{b}ar(BC)D$

c$.(B+BC)(B\frac{+}{b}ar(B)C)(B+D)$

d$.$ ABCD$+AB(\frac{?}{\text{b}\text{a}\text{r}}(CD))+(\frac{?}{\text{b}\text{a}\text{r}}(AB))CD$

e$.$ ABC$[AB\frac{+}{b}ar(C)(BC+AC)]$