finding dual means shifting from positive logic to negative logic and vice versa . hence 0 will change to 1 , 1 to 0 , and to or .
consider the table for a<=b and a^b=a
e.g
a |
b |
a^b |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
not valid because a>b |
1 |
1 |
1 |
now consider the dual of above table
a |
b |
a or b =a |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
not valid |
0 |
0 |
0 |
we can clearly see that if a or b =a then a>= b
therefore option d is correct