GATE CSE 2019 | Question: 6

Question

Which one of the following is NOT a valid identity?

• A. $(x \oplus y) \oplus z = x \oplus (y \oplus z)$
• B. $(x + y) \oplus z = x \oplus (y+z)$
• C. $x \oplus y = x+y, \text{ if } xy=0$
• D. $x \oplus y = (xy+x’y’)’$

Comments

all logic gates are commutative are associative except for nand and nor which are commutative but not associative

x xor y we think think as ~xy + x~y even if xy =0 we cant conclude anything,

but if we use like ~(xy + ~x~y) then if xy =0,

x xor y = ~(~x~y) = x+y

---

Detailed Video Explanation: https://youtu.be/DbKnUyyzlZs?t=716

---

Only Option B is valid
Answer Option B

Answer 1

1. $XOR$ is associative so $(x \oplus y) \oplus z = x \oplus (y \oplus z)$
2. For $2$ input, XOR and XNOR are complement to each other i.e. $x \oplus y = (xy + x^{'}y^{'})^{'} $
3. $x \oplus y = x+y \text{ if }xy = 0$

Only false statement is option B.

Comments

$x \oplus y = (x \odot y)' = (xy + x'y')'$

given xy = 0

=> $x \oplus y  $= (x'y')' = x+y

---

xy=0 means (x=0 and y=1) or ( x=1 and y=0) or (x=0 and y=0 ) . but C is not satisified for initial two conditions .

---

@Neelam_$ingh_222 

$x⊕y=xy'+x'y$ 

if Y=0 then  $x⊕y=x$ 

if X=0 then $x⊕y=y$ 

if X=0 , Y=0 then $x⊕y=0$

$⇒$ $x⊕y =x+y$   $if$   $xy=0$

Answer 2

$A. \:(x \oplus y) \oplus z = x \oplus (y \oplus z)$

$\Longrightarrow EX-OR$ 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 $(B).$

Answer 3

Checking option A = XOR is associative so this is valid.

Checking option C =

     xy=0 means x=y' .

     so x XOR y = x'y +xy' = (y')' y + x x = x+y

     so option c is valid.

Checking option D = (xy + x'y')' = (xy)' (x'y')' = (x'+y')(x+y) = x'x + x'y +y'x +y'y = x XOR y so option D is valid.

Checking option B = Remaining only option B is the ans and by expanding we can check it's not valid.

Answer 4

option C is true by truth table 

X	Y	X ex-or Y	x+y	xy
0	0	0	0	0
0	1	1	1	0
1	0	1	1	0
1	1	0	1	1

option A says associativity of ex-or,(so true),

option D is true because complement of ex-nor is ex-or(for even variables),

so answer is option B.