# Simplification of boolean algebra

• Jan 10th 2009, 12:25 AM
Urfreiend
Simplification of boolean algebra
(Happy)PLEASE LOOK AT THIS TABLE

(Bow)TABLE 1. POSTULATES OF BOOLEAN ALGEBRA
Basic Identity
(A) x+0 = x
(B) x . 1 = x
Commutative Property
(A) x+y = y+x
(B) xy = yx
Distributive Property
(A) x (y+z) = xy + xz
(B) x+yz = (x+y) (x+z)
Basic Identity
(A) x+x’ = 1
(B) x . x’ = 0

(Bow)TABLE 2. THEOREMS OF BOOLEAN ALGEBRA
Theorem 1: Identity
(A) x+x = x
(B) x.x = x
Theorem 2: Identity
(A) x+1 = 1
(B) x.0 = 0
Theorem 3: Involution
(x’)’= x

Theorem 4: Associative
(A) x+(y+z) = (x+y)+z
(B) x(yz) = (xy)z
Theorem 5: DeMorgan
(A) (x+y)’= x’y’
(B) (xy)’ = x’+y’
Theorem 6: Absorption
(A) x+xy = x
(B) x(x+y) = x

Now look at this sample problem and solution:

SIMPLIFY ABC+A’B’C+A’BC+ABC’+A’B’C’
(Clapping)SOLUTION
(Nerd)REASON
=ABC+A’BC+A’B’C+A’B’C’+ABC’
Commutative (a)
=BC (A+A’) + A’B’ (C+C’) + ABC’
Distributive (a)
=BC (1) + A’B’ (1) + ABC’
Basic Identity (a)
=BC+ A’B’+ABC’
Basic Identity (b)
=A’B’ + BC +ABC’
Commutative (a)
=A’B’+B (C+AC’)
Distributive (a)
=A’B’{B’+C’ (A’+C)}
DeMorgan (a,b)
=A’B’ (B’+C’A+C’C)
Distributive (a)
=A’B’ (B’+A’C+CC’)
Commutative (b)
=A’B’ (B’+A’C’+0)
Basic Identity (b)
=A’B’ (B’+A’C’)
Basic Identity (a)
=A’B’+B (A+C)
DeMorgan (a,b)

Now, can you simplify this Boolean equation using Postulates and theorems of Boolean Algebra into 3 terms only?

(Wait)WXYZ + W’XYZ + WX’YZ + WXYZ’ + W’X’YZ + WXY’Z’ + W’X’Y’Z +W’X’Y’Z’(Itwasntme)

Thanks!
• Jan 10th 2009, 01:12 AM
Isomorphism
I will suggest an easier method. Write the Karnaugh map for that expression, then simplify it to the three terms you require. Do you know how to write the Karnaugh map for logic expressions?
• Jan 10th 2009, 05:58 AM
Urfreiend
simplification of boolean algebra
What is required is to simplify it using postulates and theorem but not by using Karnaugh map... Thanks
• Jan 11th 2009, 11:20 AM
nvv
Hi, Urfreiend. And what is a problem? There is a sample, so simply do like in the sample (Happy)

Here is in my attempt. I obtain 4 terms, but maybe You'll find mistake in my solution and obtain correct solutions with 3 terms only (Itwasntme)

WXYZ + W’XYZ + WX’YZ + WXYZ’ + W’X’YZ + WXY’Z’ + W’X’Y’Z +W’X’Y’Z’ =

|
WXYZ + W’XYZ = Distributive Property = XYZ(W+W')= Basic Identity=XYZ(1) =XYZ|
|W’X’Y’Z +W’X’Y’Z’ =Distributive Property =W’X’Y’(Z+Z')= Basic Identity=W’X’Y’(1)=W’X’Y’|

=
XYZ + WX’YZ + WXYZ’ + W’X’YZ + WXY’Z’ + W’X’Y’ =

|
XYZ + WX’YZ = Distributive Property = YZ(X+WX’)=Distributive Property =YZ((X+W)(X+X'))=Basic Identity=YZ((X+W)(1))=YZ((X+W))=YZX+YZW|

=
YZX+YZW+ WXYZ’ + W’X’YZ + WXY’Z’ + W’X’Y’ =

|
YZW + W’X’YZ=(from this place I woun't write POSTULATES and THEOREMS names)=YZ(W+W’X’)=YZW+YZX'|

=
YZX+YZW+YZX'+WXYZ’ +WXY’Z’ + W’X’Y’ =

|
YZX+YZX'=YZ|

YZ+YZW+WXYZ’ +WXY’Z’ + W’X’Y’ =

|
YZW+WXYZ’=YW(Z+XZ')=YWZ+YWX|

=
YZ+YWZ+YWX+WXY’Z’ + W’X’Y’ =

|
YWX+WXY’Z’=WX(Y+Y'Z')=WXY+WXZ'|

=
YZ+YWZ+WXY+WXZ’+W’X’Y’ =

|
YZ+YWZ=YZ(1+W)=YZ(1)=YZ|

=YZ+
WXY+WXZ’+W’X’Y’ =...
• Jan 12th 2009, 01:05 AM
Urfreiend
Anyone
Anyone who can answer this problem is great! Please use the theorem and postulates to simplify it