9. apply demorgan’s theorems to each expression:\na. $overline{a + b}$\nb. $overline{overline{a}b}$\nc…

Explanation:

Step1: Recall DeMorgan's theorems

DeMorgan's theorems state that $\overline{A + B}=\overline{A}\overline{B}$ and $\overline{AB}=\overline{A}+\overline{B}$.

Step2: Solve part a

Applying the first theorem to $\overline{A + B}$, we get $\overline{A}\overline{B}$.

Step3: Solve part b

Applying the second theorem to $\overline{\overline{A}B}$, we have $A+\overline{B}$.

Step4: Solve part c

For $\overline{A + B + C}$, by extending the first theorem, we get $\overline{A}\overline{B}\overline{C}$.

Step5: Solve part d

Applying the second - theorem to $\overline{ABC}$, we have $\overline{A}+\overline{B}+\overline{C}$.

Step6: Solve part e

First expand $A(B + C)=AB+AC$. Then $\overline{A(B + C)}=\overline{AB + AC}$. By DeMorgan's theorem, $\overline{AB + AC}=\overline{AB}\overline{AC}=(\overline{A}+\overline{B})(\overline{A}+\overline{C})$.

Step7: Solve part f

For $\overline{\overline{A}B+\overline{C}D}$, by DeMorgan's theorem, we have $\overline{\overline{A}B}\overline{\overline{C}D}=(A + \overline{B})(C+\overline{D})$.

Step8: Solve part g

Applying DeMorgan's theorem to $\overline{AB + CD}$, we get $\overline{AB}\overline{CD}=(\overline{A}+\overline{B})(\overline{C}+\overline{D})$.

Step9: Solve part h

For $\overline{(A + B)(\overline{C}+D)}$, by DeMorgan's theorem, we have $\overline{A + B}+\overline{\overline{C}+D}=\overline{A}\overline{B}+C\overline{D}$.

Answer:

a. $\overline{A}\overline{B}$ b. $A+\overline{B}$ c. $\overline{A}\overline{B}\overline{C}$ d. $\overline{A}+\overline{B}+\overline{C}$ e. $(\overline{A}+\overline{B})(\overline{A}+\overline{C})$ f. $(A + \overline{B})(C+\overline{D})$ g. $(\overline{A}+\overline{B})(\overline{C}+\overline{D})$ h. $\overline{A}\overline{B}+C\overline{D}$