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’

=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!**

simplification of boolean algebra

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