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?
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=ABC+A’BC+A’B’C+A’B’C’+ABC’REASON
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!
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’ =...