Question

Simplify the following Boolean expressions to the minimum number of terms using the properties of Boolean algebra (show your work and write the property you are applying). State if they cannot be simplified.

A. A’B + AB

B. XY + X(WZ + WZ’)

C. X’Y’(X’+Y)(Y’+Y)

D. ABC + A’B + A’BC’

E. (A+B)(AC+AC’)+AB+B

Draw the circuit logic diagrams for both the original and simplified expressions.

Answer 1

A.

A'B+AB

=(A'+A)B {Distributive law}

=1*B {Anything ORed with its own complement equals 1}

=B {Anything ANDed with a 1 is equal to itself}

circuit diagram of original expression:

Circuit diagram of simplified expression is just B .

B. XY + X(WZ + WZ')

= XY + X*W*(Z+Z') {Distributive law}

= XY + X*W*1 {Anything ORed with its own complement equals 1}

=XY + XW

= X*(Y+W)

Circuit diagram of original expression:

Circuit diagram of simplified expression:

3. X'Y'(X'+Y)(Y'+Y)

=X'Y'(X'+Y)(1)        {Y'+Y = 1}

=X'Y'X' + X'Y'Y

=X'Y' + X'Y'Y

=X'Y' {Y'Y = 0, X'*0 = 0}

Circuit diagram of original expression:

Circuit diagram of simplified expression:

D. ABC + A'B + A'BC'

= ABC + A'BC' + A'B {ASSOCIATIVE LAW}

= B(AC + A'C') + A'B {DISTRIBUTIVE LAW}

= B(AC + (AC)') + A'B {(AC)' = A'C'}

= B + A'B {A+A'=1. HERE (AC)' IS COMPLEMENT OF AC}

=(1 + A')B {DISTRIBUTIVE LAW}

= B          {Anything ORed with a 1 is equal to 1. A' + 1 = 1}

Circuit diagram of original expression:

Circuit diagram of simplified expression is just B.

E. (A+B)(AC+AC')+AB+B

= (A+B)(A(C+C'))+AB+B {DISTRIBUTIVE LAW}

= (A+B)A + (A+1)B       {C+C' = 1. DISTRIBUTIVE LAW}

=AA + AB + B                 {A+1 = 1, DISTRIBUTIVE LAW}

= A + AB + B                 {AA=A}

= A + B                 {A+AB = A(1+B) = A}

Circuit diagram of original expression:

Circuit diagram of simplified expression: