# Boolean Expression + algebraic manipulation

Thread Starter

#### mnv_smiley

Joined Sep 11, 2007
3
Could some one please help me! I do not understand this algebraic manipulation and would appreciate any help.

Prove the identity of each of the following Boolean expressions, using algebraic manipulation:

a) ABC' + BC'D' + BC + C'D = B + C'd
b) WY + W'YZ' + WXZ + W'XY' = WY + W'XZ' + X'YZ' + XY'Z
c) AD' + A'B + C'D + B'C = (A'+B'+C'+D') (A + B + C + D)

any help at all would be greatly appreciated. Even if it's just a starting hint or soemthing. #### Dave

Joined Nov 17, 2003
6,970
Its probably best to give you some pointers first, then you can try and post up your answers so we can check.

Check the e-book sections:

Boolean algebraic identities

Boolean algebraic properties

Boolean rules for simplification

Pointers:

Q1) Look for common factors, and note that the CD' term is on both sides of the equation (therefore might not need touching)

Q2) The above identifies should be useful here. Note this is more of a rearrangement exercise.

Q3) Look at DeMorgans Theorems for this one.

Dave

Thread Starter

#### mnv_smiley

Joined Sep 11, 2007
3
I factored b out and got:

b(ac' + c'd' + c) + c'd = b + c'd
then since I have a c and a c' I got rid of the c and I now have:

b(a + d') + c'd = b + c'd

#### Dave

Joined Nov 17, 2003
6,970
I factored b out and got:

b(ac' + c'd' + c) + c'd = b + c'd
then since I have a c and a c' I got rid of the c and I now have:

b(a + d') + c'd = b + c'd
Can I ask by what process you got rid of the C?

Dave