EX NOR Gate

Question

$A\odot(BC) = (A\odot B)(A\odot C)$

Does it hold? Please solve.

Comments

You can check this easily using a truth table.

What's the problem?

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TAKE A=0 B=1 C=0.

A⊙(BC)=0⊙0 =1

(A⊙B)(A⊙C)= 0⊙1 * 0⊙0 = 0 *1 =0

fails .

therefore lhs <> rhs.

 

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@goxul good idea. Actually, it was supposed to be true and I though I was doing some mistake. And so just wanted to know. Tnanks!

@arvin thanks!

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I think there should be an xnor between two operations in the.rhs. to be true... I haven't checked that...check it once.

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it is simply  " Is Ex-NoR Distributive over AND Gate ? "

AND gate is distributive over OR, Ex-OR

OR gate is distributive over AND, Ex-NOR

Ex-OR gate is not distributive over ANY gate (OR,AND,Ex-NOR)

Ex-NOR gate is not distributive over ANY gate (OR,AND,Ex-OR)

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@arvin , actually that does not hold either.

@Shaik Masthan ,thanks! It was helpful.

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xnor is not distributive over ^

best way to check evaluate both l.h.s & r.h.s

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@Raghav Khajuria ,thanks!