Self Doubt

Question

As per the Boolean identity x+x=x,similarly does it hold for the following expression with XOR operation too?

((P⊛Q)⊛(P⊛Q))=P⊛Q

Comments

Basic property  of xor gate is A ex-or A =0

Therefore let assume ( P exor Q ) = A....

 

Exor follows idempotent law when you use with even no.of gates, means ( A exor A exor A ) = A

In other words....

Exor follows Idempotent law when you use With odd no.of variables

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@Shaik Masthan,So as per idempotent law what I have assumed is correct?Isn't it?

The LHS satisfies the RHS as well. :). Thanks a lot brother. :)

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brother i am not getting what you understood...

((P⊛Q)⊛(P⊛Q))=P⊛Q is false, due to you have odd no.of ex-or gates ( no.of ex-or gates are 1 but not 3 in your question ).

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A exor A exor A = A

But A exor A =0

Answer 1

Xor gate is commutative as well as associative.

 

Comments

@abhishekmehta4u,Indeed a big thanks. You've cleared my doubt. :)

Answer 2

No

let....p=0 and q=1 then l

L.H.S.=(0 EXOR 1) exor (0 exor 1)

=1 exor 1

=0

 

R.H.S = 0 exor 1

=1

Comments

@Priyanka Agrawal,Thanks a lot sister. :)

Answer 3

A xor B = AB' + A'B

Now if A and B become equal then :-

A xor B=AB' +A'B .............. Put B=A in this equation

A xor A=AA' + A'A

            =0 + 0

            =0