Question

Simplify the following Boolean expressions to a minimum number of literals

a. xy+xy’
b. (x+y)(x+y’)
c. (a+b+c’)(a’b’+c)
d. a’bc+abc’+abc+a’bc’
e. (x+y)’(x’+y’) f. xy+x(wz+wz’)

Answer 1

Answer:

a. xy+xy’  = x

b. (x+y)(x+y’)  = x

c. (a+b+c’)(a’b’+c)  = c (a + b) + a'b'c'

d. a’bc+abc’+abc+a’bc’  = b

e. (x+y)’(x’+y’) =  x'y'

f. xy+x(wz+wz’) =  x(y+w)

Explanation:

a)   xy+xy’

    = x ( y + y′ )    // taking x common

    = x . 1             // y + y' = 1

    = x

b) (x+y)(x+y’)

  =  xx + xy′ + yx + yy′     //multiplying x+y with x+y'

  = x + xy′ + xy + 0          // xx = 1    yy' = 0

  = x (1+ y′ + y)               //  taking x common

  = x.1                             // y+y' = 1

 = x

c)  (a+b+c’)(a’b’+c)

  = aa'b' + ac + a'b'b + bc + a'b'c' + cc'

  = 0 + ac + 0 + bc + a'b'c' + 0   // aa'= 0  bb'=0  cc'= 0

  = ac + bc + a'b'c'  

 = c (a + b) + a'b'c'

d) a’bc + abc’ + abc + a’bc’

  = b( a'c + ac' + ac + a'c')      //taking b common

  = b( a'c + a'c' + ac' + ac)    

  = b( a'(c+c') + a(c+c')            // taking a common

  = b(a'(1) + a(1))                        //c+c' =1

  = b(a'+a)                                 //a+a'=1

  = b(1)

  = b

e) (x+y)'(x'+y')

 = ( (x+y)'(x'+y')  )

 = ( (x+y) + (x'+y')' )'

 = ( x + y + (x''y'') )'

 = ( x + y +  (xy) )'

 = ( x +y + xy )'

 = (  x ( 1 + y ) +y  )'

 = (  x + y  )'

  = x'y'

OR it can be done as :

e) (x+y)'(x'+y')

= x′y′ (x′+y′)                 //(x+y)' = x'y'

= x′y′ + x′y′                  // (xy)' = x'+y'

= x′y′

f. xy+x(wz+wz’)

= xy + xw(z+z′)             //taking w common

= xy + xw                     // z+z'=1

= x(y+w)

Answer 2

Boolean expressions are expressions that have values of 0 or 1.

(a) xy+xy’

Rewrite as:

$\mathbf{xy+xy' = x(y + y')}$

Apply complementary law

$\mathbf{xy+xy' = x.1}$

By identity law, we have:

$\mathbf{xy+xy' = x}$

Hence, the simplified expression of xy + xy' is x

(b) (x+y)(x+y’)

Distribute the above expression

$\mathbf{(x +y)(x + y') = xx + xy' + xy + yy'}$

Simplify

$\mathbf{(x +y)(x + y') = x + xy' + xy + 0}$

$\mathbf{(x +y)(x + y') = x + xy' + xy}$

In (a), we have: $\mathbf{xy+xy' = x}$

So, the expression becomes

$\mathbf{(x +y)(x + y') = x + x}$

This gives

$\mathbf{(x +y)(x + y') = x }$

Hence, the simplified expression of (x + y)(x + y') is x

(c) (a+b+c’)(a’b’+c)

Distribute the above expression

$\mathbf{(a + b + c)(a'b' +c) = aa'b' + ac + a'b'b + bc + a'b'c' + cc'}$

Apply complementary law

$\mathbf{(a + b + c)(a'b' +c) = 0.b' + ac + a'.0 + bc + a'b'c' + 0}$

Simplify

$\mathbf{(a + b + c)(a'b' +c) = ac + bc + a'b'c'}$

Distribute ac + bc

$\mathbf{(a + b + c)(a'b' +c) = (a + b)c + a'b'c'}$

Hence, the simplified expression of (a + b + c)(a'b' +c)  is (a + b)c + a'b'c'

(d) a'bc + abc' + abc + a'bc'

Distributive the above expression

$\mathbf{a'bc + abc' + abc + a'bc' = (a'c + ac' + ac + a'c')b}$

Distribute

$\mathbf{a'bc + abc' + abc + a'bc' = ((c+c')a' + (c+c')a)b}$

Apply complementary law

$\mathbf{a'bc + abc' + abc + a'bc' = ((1)a' + (1)a)b}$

$\mathbf{a'bc + abc' + abc + a'bc' = (a' + a)b}$

Apply complementary law

$\mathbf{a'bc + abc' + abc + a'bc' = (1)b}$

Simplify

$\mathbf{a'bc + abc' + abc + a'bc' = b}$

Hence, the simplified expression of a'bc + abc' + abc + a'bc' is b

(e) (x+y)'(x'+y')

Apply De Morgan's law

$\mathbf{(x+y)'(x'+y')= x'y' (x'+y')}$

Apply De Morgan's law

$\mathbf{(x+y)'(x'+y')= x'y' + x'y'}$

Simplify

$\mathbf{(x+y)'(x'+y')= x'y'}$

Hence, the simplified expression of (x+y)'(x'+y') is x'y'

(f) xy+x(wz+wz')

Distribute

$\mathbf{xy+x(wz+wz')= xy + x(z+z')w}$

Apply complementary law

$\mathbf{xy+x(wz+wz')= xy + x(1)w}$

$\mathbf{xy+x(wz+wz')= xy + xw}$

Distribute

$\mathbf{xy+x(wz+wz')= (y + w)x}$

Hence, the simplified expression of xy + x(wz +wz') is (y + w)x

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