1. Simplify the following Boolean expressions using algebraic methods. 1. A(A+B)+B(B+C)+C(C+A) 2. (A+B ̅)(B+C)+(A+B)(C+A ̅) 3. (A+B)(AC+AC ̅)+AB+B 4. A ̅(A+B)+(B+A)(A+B ̅)

Basic Rules:

1.) $A(A+B)+B(B+C)+C(C+A)$

$\Rightarrow A\cdot A+A\cdot B+B\cdot B+B\cdot C+C\cdot C+C\cdot A\\\Rightarrow A+AB+B+BC+C+CA\ \ \{\because A\cdot A=A\}$

$\Rightarrow A(1+B)+B(1+C)+C(1+A)\\\Rightarrow A+B+C\ \ \ \{\because \ 1+A=1\}$

2.) $(A+\bar B)(B+C)+(A+B)(C+\bar A)$

$\Rightarrow AB+AC+\bar BB+\bar BC+AC+A\bar A+BC+B\bar A\\\Rightarrow AB+AC+BC+\bar BC+\bar BA\ \ \ \{\because\ A \bar A=B\bar B=0\ \ \}$

$\Rightarrow B(A+\bar A)+AC+C(B+\bar B)\\\Rightarrow B\cdot 1+AC+C\cdot1\\\Rightarrow B+C(1+A)\\\Rightarrow B+C$

3.) $(A+B)(AC+A\bar C)+AB+B$

$\implies (A+B)\cdot A(C+\bar C)+B(1+A)\\\implies A(A+B)+B\\\implies A\cdot A +AB+B\\\implies A+AB+B\\\implies A(1+B)+B\\\implies A+B$

4.) $\bar A(A+B)+(B+A)(A+\bar B)$

$\implies \bar A\cdot A+\bar A\cdot B+BA+\bar BB+AA+A\bar B\\\implies 0+\bar AB+BA+0+AA+A\bar B\ \ \ \{\because\ \ \bar A\cdot A=0\ \&\ A\cdot A=A\}$

$\implies A+BA+\bar AB+A\bar B\\\implies A(1+B)+\bar AB+A\bar B\ (\because \ 1+B=1)\\\implies A+\bar AB+A\bar B\\\implies A(1+B)+\bar AB+A\bar B\\\implies (AA+AB)+\bar AB+A\bar B\\\implies AA+AB+A\bar A+\bar AB+A\bar B\\\implies (A+\bar A)(A+B)+A\bar B\\\implies 1\cdot(A+B)+A\bar B\\\implies A+B+A\bar B\\\implies A(1+\bar B)+B\\\implies A+B$