$$ 2^{(AΔB)}=\{S|S = X\cup Y\} $$ where $ X∈ 2^A, Y∈ 2^B$, and $$S= Z\cup W $$ where $Z∈ 2^{(A)^c}, W∈2^{(B)^c}$
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What is the meaning of $\Delta$ in the set $A\Delta B$? – Mark Fischler Mar 20 '19 at 19:25
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@MarkFischler (A\cup B) - (A\cap B) – Mahmood Jazmawy Mar 20 '19 at 19:26
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symmetric difference – Aditya Prakash Mar 20 '19 at 19:28
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@Mahmood Jazmawy your question seems to be wrong , $S=X \cup Y -(X \cap Y)$ , according to me this should be equivalent to $2^{A \Delta B}$ , should this be true , then I can post an answer , please verify. – Aditya Prakash Mar 20 '19 at 19:30
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@ADITYAPRAKASH no the question as is I didn't change anything in it and the 2AΔB is as I said – Mahmood Jazmawy Mar 20 '19 at 19:35
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@Mahmood Jazmawy , read my answer , I have stated why your question is probably wrong too. – Aditya Prakash Mar 20 '19 at 19:59
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I thought a diagram will be useful here.As you can see I have drawn sets A & B and X & Y.The symmetric difference of A and B i.e. $A\Delta B$ is coloured in buff color(light orange).You have defined the process of moving from set $A\cup B$ to $X\cup Y$ as raising 2 to the power of elements from the former to the latter.It's evident that you cannot select such a power of 2 from the set $A\cup B$ that lies in $A\cap B$.It implies seamlessly that the valid elements in set S won't fall in $X\cap Y$ because you are not able to select elements from $A\cap B$ in the first place and thus $$2^{(A\Delta B)}=\{S|S=(X\cup Y)-(X\cap Y)\}$$. This is the first part of your answer , You can do the second part in a similar fashion.
Aditya Prakash
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