0
$\begingroup$

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

  • 0
    I actually used "group" from the word Menge (german), although the word קבוצה is in fact the word I'm looking for (having studied this subject in hebrew as well)2011-11-08

2 Answers 2

2

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

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

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

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

1

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}