Can someone help me prove the following?

(A XOR B)' = A' XOR B

Please?

 

Hello,

Have a look at this page of the eBook:
http://www.allaboutcircuits.com/vol_4/chpt_3/4.html

You can also make a state table and compare.

Bertus

 

Hello thisonedude

I think it would be a lot of words trying to prove that (A XOR B)' = A' XOR B.

Maybe if you look at the attached images you can find it.
In these cases it would be faster to search in Google.com for answers.

Among the many answers I found is that you look at the following link:
http://pubs.sciepub.com/jcsa/1/1/3/
I hope you serve

 

(a xor b)' = (a xor b) xor 1 = a xor b xor 1 = a xor 1 xor b = (a xor 1) xor b = a' xor b.

Edit: It was all in capitals. Forum software has changed it.

 

You ca prove it either by making a table, or using boolean algebra.

 

\(

(a'b + ab)' = [(a'b)'][(ab')]

(a'' + b')(a' + b'') = (aa') + (ab) + (a'b') + (bb') = a' xor b

\)

 

A XOR B is an inequality function.
Hence (A XOR B)' - is equality function

Hence (A XOR B) ' = (A' XOR B) = (A XOR B')

This is not a proof but rationalized logic.