# Let A be nonepty set and P(A) be the power set of A. Recall the definition of power set: P(A)={x|x sube A} Show that symmetric deference operation on

Let A be nonepty set and P(A) be the power set of A. Recall the definition of power set:
$P\left(A\right)=\left\{x\mid x\subseteq A\right\}$
Show that symmetric deference operation on P(A) define by the formula
$x\oplus y=\left(x\cap {y}^{c}\right)\cup \left(y\cap {x}^{c}\right),x\in P\left(A\right),y\in p\left(A\right)$
(where ${y}^{c}$ is the complement of y) the following statement istrue:
The algebraic operation o+ is commutative and associative on P(A).
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Derrick
Commutative:
For $x\in P\left(A\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y\in P\left(A\right),$
$x\oplus y=\left(x\cap {y}^{c}\right)\cup \left(y\cap {x}^{c}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}$
$y\oplus x=\left(y\cap {x}^{c}\right)\cup \left(x\cap {y}^{c}\right)$
$=\left(x\cap {y}^{c}\right)\cup \left(y\cap {x}^{c}\right)$
$=x\oplus y$
Associative
For $x,y\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}z\in P\left(A\right)$
$\left(x\oplus y\right)\oplus z=\left(\left(x\oplus y\right)\cap {z}^{c}\right)\cup \left(zn{\left(x\oplus y\right)}^{c}\right)$
$=\left(z\cap \left(\left(x\cap {y}^{c}\right)\cup {\left(y\cap {x}^{c}\right)}^{c}\right)\cup \left(\left(\left(x\cap {y}^{c}\right)\cup \left(y\cap {x}^{c}\right)\right)\cap {z}^{c}\right)$
$=\left(z\cap \left({x}^{c}\cup y\right)\cap \left({y}^{c}\cup x\right)\right)\cup \left(\left(x\cap {y}^{c}\right)\cap {z}^{c}\right)\cup \left(\left(y\cap {x}^{c}\right)\cap {z}^{c}\right)$
$=\left(z\cap \left({x}^{c}\cap \left({y}^{c}\cup x\right)\right)\cup \left(\left(y\cap {y}^{c}\right)\cup x\right)\right)\cup \left(\left(\left(x\cap {y}^{c}\right)\cup \left(y\cap {x}^{c}\right)\right)\cap {z}^{c}\right)$
$=\left(z\cap \left(\left({x}^{c}\cap {y}^{c}\right)\cup \left({x}^{c}\cap x\right)\right)\cup \left(\left(y\cap {y}^{c}\right)\cup \left(y\cap x\right)\right)\right)\cup \left(\left(\left(x\cap {y}^{}\right)\cap {z}^{c}\right)\cup \left(\left(y\cap {x}^{c}\right)\cap {z}^{c}\right)\right)$
$=\left(z\cap \left(\left({x}^{c}\cap {y}^{c}\right)\cup \left({x}^{c}\cap x\right)\right)u\left(\left(y\cap {y}^{c}\right)\cup \left(y\cap x\right)\right)\right)\cup \left(x\cap {y}^{c}\cap {z}^{c}\right)\cup \left(y\cap {x}^{c}\cap {z}^{c}\right)$