Exercise 1. Apply De-Morgan's theorem to these Boolean expressions i. \overline{(A + B)} ii. \overline{(A + B + C)} iii. \overline{(AB + CD)} iv. \overline{(A + \overline{B}).(\overline{C} + D)} v. \overline{(P(Q + R))} 2. Simplify the Boolean expression using laws of Boolean algebra. I. A(A + B) II. \overline{A}BC + A\overline{B}C + A\overline{B}C III. AB + (\overline{A} + \overline{B})C + \overline{A}B IV. BD + B(D + E) + \overline{D}(D + F) V. ABCD + AB(\overline{CD}) + (\overline{AB})CD
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i. \(\overline{(A + B)}\) - De Morgan's theorem: \(\overline{A + B} = \overline{A} \cdot \overline{B}\) ii. \(\overline{(A + B + C)}\) - De Morgan's theorem: \(\overline{A + B + C} = \overline{A} \cdot \overline{B} \cdot \overline{C}\) iii. Show more…
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Simplify the Boolean expression $(\overline{A \cdot \bar{B}+C}) \cdot(\bar{A}+\overline{B \cdot \bar{C}})$ by using de Morgan's laws and the rules of Boolean algebra. pplying de Morgan's laws to the first term gives: $$ \begin{aligned} \overline{A \cdot \bar{B}+C} &=\overline{A \cdot \bar{B}} \cdot \bar{C}=(\bar{A}+\overline{\bar{B}}) \cdot \bar{C} \\ &=(\bar{A}+B) \cdot \bar{C}=\bar{A} \cdot \bar{C}+B \cdot \bar{C} \end{aligned} $$ pplying de Morgan's law to the second term gives: $$ \bar{A}+\overline{B \cdot \bar{C}}=\bar{A}+(\bar{B}+\overline{\bar{C}})=\bar{A}+(\bar{B}+C) $$ hus $(\overline{A \cdot \bar{B}+C}) \cdot(\bar{A}+\overline{B \cdot \bar{C}})$ $$ \begin{aligned} &=(\bar{A} \cdot \bar{C}+B \cdot \bar{C}) \cdot(\bar{A}+\bar{B}+C) \\ &=\bar{A} \cdot \bar{A} \cdot \bar{C}+\bar{A} \cdot \bar{B} \cdot \bar{C}+\bar{A} \cdot \bar{C} \cdot C \\ &\quad+\bar{A} \cdot B \cdot \bar{C}+B \cdot \bar{B} \cdot \bar{C}+B \cdot \bar{C} \cdot C \end{aligned} $$ But from Table $11,7, \bar{A} \cdot \bar{A}=\bar{A}$ and $\bar{C} \cdot C=B \cdot \bar{B}=0$ Hence the Boolean expression becomes: $$ \begin{aligned} \bar{A} & \cdot \bar{C}+\bar{A} \cdot \bar{B} \cdot \bar{C}+\bar{A} \cdot B \cdot \bar{C} \\ &=\bar{A} \cdot \bar{C}(1+\bar{B}+B) \\ &=\bar{A} \cdot \bar{C}(1+B) \\ &=\bar{A} \cdot \bar{C} \end{aligned} $$ Thus: $\overline{(A \cdot \bar{B}+C}) \cdot(\bar{A}+\overline{B \cdot \bar{C}})=\bar{A} \cdot \bar{C}$
Simplify the Boolean expression $(\overline{\bar{A} \cdot B})+(\overline{\bar{A}+B})$ by using de Morgan's laws and the rules of Boolean algebra. Applying de Morgan's law to the first term gives: $\overline{\bar{A} \cdot B}=\overline{\bar{A}}+\bar{B}=A+\bar{B}$ since $\overline{\bar{A}}=A$ Applying de Morgan's law to the second term gives: $$ \overline{\bar{A}+B}=\overline{\bar{A}} \cdot \bar{B}=A \cdot \bar{B} $$ Thus, $(\overline{\bar{A} \cdot B})+(\overline{\bar{A}+B})=(A+\bar{B})+A \cdot \bar{B}$ Removing the bracket and reordering gives: $A+A$ : $\bar{B}+\bar{B}$ But, by rule 15 , Table $11.7, A+A \cdot B=A$, It follows that: $A+A \cdot \bar{B}=A$ Thus: $(\overline{A \cdot B})+(\overline{\bar{A}+B})=A+\bar{B}$
Using boolean rules, deduce: AB + ABC + A'BC + AB'C
Adi S.
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