How do you simplify these four boolean expressions step by step?

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

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

3. Y = (C + D)' + A'CD' + AB'C' + A'B'CD + ACD'

4. Y = AB(C'D)' + A'BD + B'C'D'

 

What are the associative, distributive, and commutative properties of complement with respect to + and * in boolean logic? Your textbook has a table somewhere.

You know that A+A' = 1 and A*A'=0 for both states of A, start applying these identities to those equations.

 

Heres is my answer.. i might be wrong in some steps but still hope it will help you out. i got Q1,2&4. Q3 got confused by the "bar". here:

1) Y = A'B'C' + A'BC + ABC + AB'C' + AB'C
= A'(B'C'+BC)+ABC+AB'(C'+C)
= A'(1)+ABC+AB'(1)
= A'+ABC+AB'
=A'+BC+B'
=A'+B'+C

2) Y = (B + C')(B') + (A' + B + C')'
=B'B+B'C'+AB'C
=0+B'C'+AB'C
=B(C'+AC)
=B(C'+A)
=BC'+AB

3)DUNNOE
4) Y = AB(C'D)' + A'BD + B'C'D'
=AB(C+D')+A'BD+B'C'D'
=ABC+ABD'+A'BD+B'C'D'
=ABC+B(AD'+A'D)+B'C'D'
=ABC+B(1)+B'C'D'
=ABC+B+B'C'D'
=ABC+B+C'D'

 

Originally posted by ecjohnny@Feb 16 2006, 10:34 AM

3)DUNNOE

[post=14079]Quoted post[/post]​

The first simplification on this one is that
A'CD' + ACD' = CD'
and the second simplification is that
( C + D)' + CD' = D'
So we are left with
D' + AB'C' + A'B'CD

 

AB'+C how can you draw this with the least number of two-input nor gates
If you know how email me the answer at babolar4444@yahoo.com

 

AB'+C how can you draw this with the least number of two-input nor gates

let me try.
just implement as usual using NOT,AND and OR gate.
adding NOT before AND will give you NOR and NOT after OR will give you NOR.
Giving the same input to both terminals of NOR would invert the function (i.e) would act as NOT as (x+x)' = x'.
I think 4 NOR gates would be required in this case