$A. \:(x \oplus y) \oplus z = x \oplus (y \oplus z)$
$\Longrightarrow EX-OR$ and $EX-NOR$ $GATE$ are Commutative and associative.
$\Longrightarrow NAND$ and $NOR$ $GATE$ are commutative but not associative.
$B.\:(x + y) \oplus z = x \oplus (y+z)$
$LHS:(x + y) \oplus z=(x+y)z'+(x+y)'z=xz'+yz'+x'.y'.z$
$RHS:x \oplus (y+z)=x(y+z)'+x'(y+z)=x.y'.z'+x'y+x'z$
Clearly $LHS\neq RHS$
So, this is not valid.
$C.\:x \oplus y = x+y, \text{if}\: xy=0$
It is like $Q\:\text{if}\:P$ means $\text{if}\: P\:\text{then}\:Q$
We can write in ${\color{Magenta}{P\rightarrow Q} }$
Now if ${\color{Red} {xy=0}},$ then ${\color{Green} {x \oplus y = x+y}}$
$x$ |
$y$ |
$xy$ |
$x\oplus y$ |
$x+y$ |
$0$ |
$0$ |
${\color{Red} 0}$ |
${\color{Green} 0}$ |
${\color{Green} 0}$ |
$0$ |
$1$ |
${\color{Red} 0}$ |
${\color{Green} 1}$ |
${\color{Green} 1}$ |
$1$ |
$0$ |
${\color{Red} 0}$ |
${\color{Green} 1}$ |
${\color{Green} 1}$ |
$1$ |
$1$ |
$1$ |
$0$ |
$1$ |
${\color{Orange}{(OR)}}$
if ${\color{Red} {xy=0}},$ then ${\color{Green} {x \oplus y = x+y}}$
We know that ${\color{DarkBlue}{x\oplus y=xy'+x'y}}={\color{Orange}{(x\odot y)'}}={\color{Green}{(xy+x'y')'}}={\color{Green} {(0+x'y')'}}={\color{Magenta}{x+y}}$
$D.\:{\color{Purple}{x \oplus y = (xy+x’y’)’}}$
So, the correct answer is $(B).$