Question
Show that
a) Using truth tables
b) Using Boolean identities
Answer
9 a) show that xz=(x+y)(x+y’)(x’+z) by using truth table.
|
x |
y |
z |
Y’ |
X’ |
xz |
X+y |
X+y’ |
X’+z |
(x+y)(x+y’)(x’+z) |
|
0 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
|
0 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
|
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
|
0 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
|
1 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
|
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
|
1 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
|
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
Hence proved that xz=(x+y)(x+y’)(x’+z)
b) show that xz=(x+y)(x+y’)(x’+z) by using Boolean algebra identities :
xz=(x+y)(x+y’)(x’+z) {First we multiply the two terms (x+y)(x+y’)}
= (x+xy’+xy+y.y’)(x’+z) {identity :- x.x’=0}
= (x+xy’+xy+0)(x’+z)
= (x+xy’+xy)(x’+z) {Now we multiply these two terms}
= x’.x+xy’.x’+xy.x’+zx+zxy’+zxy
= 0+0+0+zx+zx(y+y’) {identity :- y+y’=1}
= zx+zx {identity :- a+a=a}
= zx Hence proved that xz=(x+y)(x+y’)(x’+z)