in Digital Logic
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Show that the Dual of the Exclusive OR is equal to its complement.
in Digital Logic
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a exor b= a'b + ab'

dual of a exor b = (a'+b).(a+b')  // remember dual does not change nature of variable

                          =a'.a + a'b' +ba+ bb' =a'b' +ab = a exnor b

now, (a exor b)' = (a'b +ab')'

                         =(a+b')(a'+b)= a'b' +ab =a exnor b

hence, exor is an orthogonal function whose dual and complement both are same i.e exnor

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The so called “Orthogonal Function” is a made-up word by some YouTubers, who provide Wrong analysis of those functions whose complement & dual are same.

Watch This: Misconception in Digital Logic | So Called "Orthogonal Boolean Function"

The number of functions whose complement & dual are same is $2^{(2^{n-1})}.$ Detailed Explanation & Analysis is HERE. 

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Dual & Complement of Ex-OR is Ex-NOR. So, the Dual of the Exclusive OR is equal to its complement.

Functions whose Complement is same as their dual: Complete , Detailed Analysis with Proof 

Watch this to know the Misconceptions about such functions: Misconception in Digital Logic | So Called "Orthogonal Boolean Function"

The number of functions whose complement & dual are same is $2^{(2^{n-1})}.$ Detailed Explanation & Analysis is HERE.  

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