Simplify Boolean expression cabc' + ac'b + ba'c + c'ba'b + bb' + acb + abb.
Added by Victor L.
Step 1
Simplify Boolean expression cabc' + ac'b + ba'c + c'ba'b + bb' + acb + abb: First, we can simplify bb' to 0 (since b and its complement cannot both be true at the same time). Then, we can use the distributive property to group terms with common factors: cabc' + 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: $\overline{(A \cdot \bar{B}+C}) \cdot(\bar{A}+\overline{B \cdot \bar{C}})=\bar{A} \cdot \bar{C}$
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