Am I solving this question correctly? How can I evaluate the following term:
$$\left((\{a,b\}\cup\{b,a\})\times(\{b,a\}\cap\{a,b\})\right)\setminus
\left((\{b,a\}\setminus\{a,b\})\cup(\{a,b\}\times\{b,a\})\right)$$
You can see the notes to my approach in this picture. Am I solving it correctly?

 A: Under the assumption that you wrote the term down correctly:
$$
\begin{align}
&\left((\{a,b\}\cup\{b,a\})\times(\{b,a\}\cap\{a,b\})\right)\setminus
\left((\{b,a\}\setminus\{a,b\})\cup(\{a,b\}\times\{b,a\})\right)\\
&= \left((\{a,b\}\cup\{a,b\})\times(\{a,b\}\cap\{a,b\})\right)\setminus
\left(\emptyset\cup(\{a,b\}\times\{a,b\})\right)\\
&= \left(\{a,b\}\times\{a,b\}\right)\setminus
\left(\{a,b\}\times\{a,b\}\right)\\
&= \emptyset
\end{align}
$$
I am meanly using the folloving theorems:
$$\{a,b\}=\{b,a\}$$
$$A\cup A=A=A\cap A$$
$$A\setminus A = \emptyset$$
EDIT:
If you really need to calculate the product, just do so in the last step:
$$\begin{align}
&(\{a,b\}\times\{a,b\})\setminus(\{a,b\}\times\{a,b\})\\
&= \{(a, a),(a,b),(b,a),(b,b)\}\setminus\{(a, a),(a,b),(b,a),(b,b)\}\\
&=\{\}
\end{align}$$
LATER EDIT:
As I understood what you did in the photo, I can say that your approach is correct except one little thing:
$$\{\}\cup\{a,b,\cdots\}\neq\{\{\},a,b,\cdots\}$$
but rather $$\{\}\cup A=A$$
For all sets A. You confused this with
$$\{\{\}\}\cup\{a,b,\cdots\}=\{\{\},a,b,\cdots\}$$
Which is a mistake that often occurs.
