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!