Boolean Algebra Questions

Question1

Using Boolean algebra techniques, simplify the expression X . Y + X (Y + Z) + Y (Y + Z)

Solution:

Given: X . Y + X (Y + Z) + Y (Y + Z).

Applying distributive property, we get

X . Y + X (Y + Z) + Y (Y + Z) = X . Y + X . Y + X . Z + Y . Y + Y . Z

We know B . B = B

= X . Y + X . Y + X . Z + Y + Y . Z

We know A . B + A . B = A . B

= X . Y + X . Z + Y + Y . Z

= X . Y + X . Z + Y  [We know (B + BC = B)]

= Y + XZ      
                               

Question 2: 

XY¯XY¯ (X¯X¯ + YY) (X¯X¯ + XX)

Solution:

Given: XY¯(X¯+Y)(X¯+X)XY¯(X¯+Y)(X¯+X)

= XY¯(X¯+Y)XY¯(X¯+Y)

= (X¯+Y¯)(X¯+Y)X¯+Y¯)(X¯+Y)

= X¯+Y.Y¯X¯+Y.Y¯

= X¯X¯

Question 3 

Simplify X+YX¯X+YX¯

Solution:

Given: XX + YX¯YX¯

By using DeMorgan’s law, AB¯AB¯ = A¯+B¯A¯+B¯

X+YX¯=X+(Y¯+X¯)X+YX¯=X+(Y¯+X¯)

= (X+X¯)+Y¯(X+X¯)+Y¯

= 1+Y¯