Problem-48
Use a table to express the values of each of these Boolean functions.
a) F(x,y,z)= 
b) F(x,y,z)=y+xz
c) F(x,y,z)=x 
+
d) F(x,y,z)=y(xz+ 
)
Solution
a) F(x,y,z)=
|
x |
y |
z |
|
|
1 |
1 |
1 |
0 |
|
1 |
1 |
0 |
0 |
|
1 |
0 |
1 |
0 |
|
1 |
0 |
0 |
0 |
|
0 |
1 |
1 |
1 |
|
0 |
1 |
0 |
1 |
|
0 |
0 |
1 |
0 |
|
0 |
0 |
0 |
0 |
b) F(x,y,z)=y+xz
|
x |
y |
z |
y+xz |
|
1 |
1 |
1 |
1 |
|
1 |
1 |
0 |
1 |
|
1 |
0 |
1 |
1 |
|
1 |
0 |
0 |
0 |
|
0 |
1 |
1 |
1 |
|
0 |
1 |
0 |
1 |
|
0 |
0 |
1 |
0 |
|
0 |
0 |
0 |
0 |
c) F(x,y,z)=x +
|
x |
y |
z |
x |
|
1 |
1 |
1 |
0 |
|
1 |
1 |
0 |
1 |
|
1 |
0 |
1 |
1 |
|
1 |
0 |
0 |
1 |
|
0 |
1 |
1 |
1 |
|
0 |
1 |
0 |
1 |
|
0 |
0 |
1 |
1 |
|
0 |
0 |
0 |
1 |
d) F(x,y,z)=y(xz+ )
|
x |
y |
z |
y(xz+ |
|
1 |
1 |
1 |
1 |
|
1 |
1 |
0 |
0 |
|
1 |
0 |
1 |
0 |
|
1 |
0 |
0 |
0 |
|
0 |
1 |
1 |
0 |
|
0 |
1 |
0 |
1 |
|
0 |
0 |
1 |
0 |
|
0 |
0 |
0 |
0 |