Proving a Discrete Mathematics equation

I'm trying to prove the following 2 equations:
A,B are subsets of a group U

A ⊕ ∅ = A
now as I understood this equation means {a,∅},{b,∅}... if a,b -> A
Right?

Second equation is

(A ⊕ B) ⊕ B = A

I think if I would understand the method on one of them,
I'll understand it on the other

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Do you have a definition of the $\oplus$ symbol? –  Henning Makholm Nov 5 '11 at 18:00
The question is unclear: your first equation is missing $B$ and you haven't defined the circled plus, which you should do for a homework question. If you're having trouble with math formatting and don't know LaTeX, I suggest writing your question out in words. This post How do I type math in my question/answer/comment? was helpful for me--it's from the FAQ. –  sasha Nov 5 '11 at 18:01
Also, are you working with groups (an algebraic structure) or simply with sets? The two questions make sense if $A\oplus B$ means the symmetric difference $(A\cup B)\setminus(A\cap B)$, but your "$\{a,\varnothing\},\{b,\varnothing\},\ldots$" could suggest that you're confusing it with direct sums. –  Henning Makholm Nov 5 '11 at 18:06
Asaf, if you want to write the name of $\oplus$ in Hebrew and I could translate it to the proper term. Do you mean הפרש סימטרי, also known as symmetric difference and "xor"? –  Asaf Karagila Nov 5 '11 at 18:10
I also note that you used "group" which can be the naive translation of קבוצה, however it is not the correct term. Group is a distinct concept translated as חבורה. –  Asaf Karagila Nov 5 '11 at 21:24

If I remember correctly, $A⊕B=(A\setminus B) \cup ( B \setminus A).$

Well, let's use it on your examples :)

$$A⊕\varnothing=(A\setminus \varnothing) \cup (\varnothing \setminus A)=A\cup\varnothing=A$$

$$(A⊕B)⊕B=((A\setminus B) \cup ( B \setminus A))⊕B=$$ $$(((A\setminus B) \cup ( B \setminus A))\setminus B)\cup(B \setminus((A\setminus B) \cup ( B \setminus A)))=$$ $$(A \setminus B)\cup(A\cap B)=A$$

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The simplest definition I know of symmetric set difference is $$x \in A \oplus B \;\equiv\; x \in A \not\equiv x \in B$$ Using this, we can calculate the elements of the left hand side of first equality, as follows: \begin{align} & x \in A \oplus \varnothing \\ \equiv & \;\;\;\;\;\text{"definition of $\;\oplus\;$"} \\ & x \in A \not\equiv x \in \varnothing \\ \equiv & \;\;\;\;\;\text{"definition of $\;\varnothing\;$"} \\ & x \in A \not\equiv \text{false} \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & x \in A \\ \end{align} By set extensionality this proves the first equality.
To prove the second, we calculate similarly \begin{align} & x \in (A \oplus B) \oplus B \\ \equiv & \;\;\;\;\;\text{"definition of $\;\oplus\;$, twice; drop parentheses since $\;\not\equiv\;$ is associative"} \\ & x \in A \not\equiv x \in B \not\equiv x \in B \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & x \in A \not\equiv \text{false} \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & x \in A \\ \end{align}