Thread: Simplification of boolean algebra

PLEASE LOOK AT THIS TABLE


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

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’

SOLUTION

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?

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


Thanks!

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?

What is required is to simplify it using postulates and theorem but not by using Karnaugh map... Thanks

Hi, Urfreiend. And what is a problem? There is a sample, so simply do like in the sample

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

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’ =...

Anyone who can answer this problem is great! Please use the theorem and postulates to simplify it