Siva Tarun You have $4(2^{2}=4)$combinations in the truth table.
All these combinations can either take value 0 or 1.
p |
q |
f0 |
f1 |
f2 |
f3 |
f4 |
f5 |
f6 |
f7 |
f8 |
f9 |
f10 |
f11 |
f12 |
f13 |
f14 |
f15 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
f0=0, f1= x.y, f2=x.y', f3= x, f4=x'.y, f5=y, f6=x xor y, f7= x +y, f8=(x+y)' , f9= x xnor y, f10=y', f11=x+y', f12 = x', f13= x'+y, f14=(x.y)', f15=1
Number of functions = 16