Distributivity Theorem in Boolean Algebra

[SpaceDomain]

Hello. I am wondering if the distributivity theorem works for factoring out more than a single term.

Distributivity Theorem:
$$BC + BD = B(C + D)$$

But can I do this:
[ tex ] ABC + ABD = AB(C + D) [ /tex ]

Thanks.

 

[micromass]

Yes, essentially your just using X(Y+Z)=XY+XZ with X=AB, Y=C and Z=D...

 

[SpaceDomain]

Awesome. Thanks.

 

[SpaceDomain]

Okay, so I am using this concept in the following problem but am getting stuck.

I am trying to simplify the following expression (A' is the complement of A) :

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

Y= A'B'(C' + C) + AB'C + A'BC + ABC

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

Y = A'B' + AB'C + BC(A + A')

Y = A'B' + AB'C + BC

Y = A'B' + C(AB' + B)

Y = A'B' + C(A + B)

Y = A'B' + CA + CBBut when I work out a Karnaugh Map I get Y = A'B' + C

How does CA + CB reduce to C?

 

[micromass]

Okay, so I am using this concept in the following problem but am getting stuck.

I am trying to simplify the following expression (A' is the complement of A) :

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

Y= A'B'(C' + C) + AB'C + A'BC + ABC

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

Y = A'B' + AB'C + BC(A + A')

Y = A'B' + AB'C + BC

Stop here. Now continue as

Y=A'B'+A'B'C+AB'C+BC

 

[SpaceDomain]

Okay, I'll try that.

But is there a way to simplify:
A'B' + CA + CB

to A'B' + C?

Or can you just get stuck when simplifying Boolean expressions?

Because when I work out the K-map on:
A'B' + CA + CB

I end up with:
A'B' + C.