Question

Use boolean algebra to prove that:

(A^- *B*C^-) + (A^- *B*C) + (A* B^- *C) + (A*B* C^-) + (A*B*C)= (A+B)*(B+C)

A^- is same as "not A"

please show steps to getting the left side to equal the right side, use boolean algebra properties such as distributive, absorption,etc

Answer 1

(A^- *B*C^-) + (A^- *B*C) + (A* B^- *C) + (A*B* C^-) + (A*B*C)

let A^- = A'

A'BC' + A'BC + AB'C + ABC' + ABC =

= A'B(C'+C) + AB'C + AB(C+C')   since C+C' = 1.   Boolean Complement Rule

= A'B + AB'C + AB

= B(A'+A) + AB'C   since A'+A = 1   Boolean Complement Rule

= B + AB'C     using rule A+BC = (A+B) (A+C)

= (B+A)(B+B'C)      using rule A+BC = (A+B) (A+C)

= (B+A)(B+B')(B+C) since B+B' = 1   Boolean Complement Rule

= (B+A)(B+C)

= (A+B)(B+C)