We use Xor Sign ->$\oplus$
We use Xnor sign->$\odot$
We have to confirm the associativity holds or not which mean we have to check ,
$\left ( \left ( a\oplus b \right )\odot c\right )$ =$\left ( a\oplus \left ( b\odot c \right ) \right )$ this holds not.
now we can confirm this by truth table.
Truth table for $\left ( \left ( a\oplus b \right )\odot c\right )$
| $a$ |
$b$ |
$c$ |
$a\oplus b$ |
$\left ( \left ( a\oplus b \right )\odot c\right )$ |
| 0 |
0 |
0 |
0 |
1 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
1 |
0 |
1 |
0 |
| 0 |
1 |
1 |
1 |
1 |
| 1 |
0 |
0 |
1 |
0 |
| 1 |
0 |
1 |
1 |
1 |
| 1 |
1 |
0 |
0 |
1 |
| 1 |
1 |
1 |
0 |
0 |
Truth table for $\left ( a\oplus \left ( b\odot c \right ) \right )$
| $a$ |
$b$ |
$c$ |
$b\odot c$ |
$\left ( a\oplus \left ( b\odot c \right ) \right )$ |
| 0 |
0 |
0 |
1 |
1 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
| 0 |
1 |
1 |
1 |
1 |
| 1 |
0 |
0 |
1 |
0 |
| 1 |
0 |
1 |
0 |
1 |
| 1 |
1 |
0 |
0 |
1 |
| 1 |
1 |
1 |
1 |
0 |
So it does not matter to which operator we associated b with $\odot$ or $\oplus$ the result of the boolean expression comes same for both the cases and it produce output 1 if the input is has even number of ones.
