3. Using Boolean Algebra laws and rules, simplify following expressions: (12 Points) a. $\overline{A}B + \overline{A}BC + \overline{A}BCD + \overline{A}BCDE$ b. $\overline{A}BC + \overline{A}B\overline{C} + \overline{A}\overline{B}C$ c. $\overline{(A + B + C + \overline{D})} + \overline{A}BCD$
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AB + ABC + ABCD + ABCDE - We can factor out AB from the first two terms: AB(1 + C + CD + CDE) - Then, we can factor out ABC from the last two terms: AB(1 + C + CD + CDE + DE) - Finally, we can factor out ABCD from the last term: AB(1 + C + CD + CDE + DE + 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: $\overline{(A \cdot \bar{B}+C}) \cdot(\bar{A}+\overline{B \cdot \bar{C}})=\bar{A} \cdot \bar{C}$
Using Boolean's law or identities, what is the most simplified form of AB + ABC + ABCD + ABCDE + ABCDEF?
Sanchit J.
Using boolean rules, deduce: AB + ABC + A'BC + AB'C
Adi S.
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