Answer to Question #259573 in Discrete Mathematics for Thuy

Question

Find the simplest form of given Boolean expressions using algebraic methods.

i. A(A+B) + B(B+C) + C(C+A)

ii. (A+|B|)(B+C) + (A+B)(C+|A|)

iii. (A+B)(AC+A|C|)_+AB +B

iv. |A|(A+B) + (B+A)(A+|B|)

Answer 1

$(i)) A(A+B)+B(B+C)+C(C+A) \\\Rightarrow \quad A \cdot A+A \cdot B+B \cdot B+B\cdot C+C \cdot C+C \cdot A \\\Rightarrow \quad A+A B+B+B C+C+C A [\because A.A=A, B.B=B, C.C=C] \\\Rightarrow \quad A(1+B)+B(1+C)+C(1+A) \\\Rightarrow \quad A+B+C \\ii) (A+\bar{B})(B+C)+(A+B)(C+\bar{A}) \\\Rightarrow A B+A C+\bar{B} B+\bar{B} C+A C+A \bar{A}+B C+B\bar{A} \\\Rightarrow \quad A B+A C+O+\bar{B }C+A C+O+B C+B \bar{A} \\\Rightarrow \quad A B+\bar{B} C+A C+A C+B C+B \bar{A} \\\Rightarrow \quad A B+\bar{B }C+A C+B C+B \bar{A}\quad[A C+A C=A C] \\\Rightarrow \quad A B+B \bar{A}+A C+\bar{B} C+B C \\\Rightarrow \quad B(A+\bar{A})+A C+C(\bar{B}+B) \\\Rightarrow \quad B+A C+C \quad[\bar{A}+A=1] \\\Rightarrow B+C(1+A) \quad[C(1+A)=C] \\\Rightarrow \quad B+C \\iii) \\(A+B)(A C+A \bar{C})+A B+B \\\Rightarrow \quad(A+B):(A(C+\bar{C}))+A B+B \\\Rightarrow \quad(A+B) A+A B+B \ [\because C+\bar{C}=1] \\\Rightarrow \quad A \cdot A+B \cdot A+A B+B \\\Rightarrow \quad A+B A+A B+B \quad[A.A=A] \\\Rightarrow \quad A(1+B)+B(1+A) \\\Rightarrow \quad A+B \quad\left[\begin{array}{c}\therefore A(1+B)=A \\ B(1+A)=B\end{array}\right] \\ iv) \\\bar{A}(A+B)+(B+A) \cdot(A+\bar{B}) \\\Rightarrow \quad \bar{A} A+\bar{A }B+B A+B \bar{B}+A \cdot A+A \bar{B} \\\Rightarrow \quad 0+\bar{A} B+B A+O+A+A \bar{B} \quad[\bar{A }A=0 ; B \bar{B}=0] \\\Rightarrow \quad B(\bar{A}+A)+A(1+\bar{B}) \\\Rightarrow \quad B+A \quad\left[\begin{array}{l}\because \bar{A}+A=1 \\ A(1+\bar{B})=A\end{array}\right]$