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In the equation, a(xor)b(xnor)c, does b belong to xor or xnor?  that equation detects even one’s !?

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.