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 P(A) define by the formula x o+ y=(x nn y^c) uu (y nn x^c), x in P(A), y in p(A) (where y^c is the complement of y) the following statement istrue: The algebraic operation o+ is commutative and associative on P(A).

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 P(A) define by the formula x o+ y=(x nn y^c) uu (y nn x^c), x in P(A), y in p(A) (where y^c is the complement of y) the following statement istrue: The algebraic operation o+ is commutative and associative on P(A).

Question
Commutative Algebra
asked 2021-02-05
Let A be nonepty set and P(A) be the power set of A. Recall the definition of power set:
\(\displaystyle{P}{\left({A}\right)}={\left\lbrace{x}{\mid}{x}\subseteq{A}\right\rbrace}\)
Show that symmetric deference operation on P(A) define by the formula
\(\displaystyle{x}\oplus{y}={\left({x}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\right)},{x}\in{P}{\left({A}\right)},{y}\in{p}{\left({A}\right)}\)
(where \(\displaystyle{y}^{{c}}\) is the complement of y) the following statement istrue:
The algebraic operation o+ is commutative and associative on P(A).

Answers (1)

2021-02-06
Commutative:
For \(\displaystyle{x}\in{P}{\left({A}\right)}{\quad\text{and}\quad}{y}\in{P}{\left({A}\right)},\)
\(\displaystyle{x}\oplus{y}={\left({x}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\right)}{\quad\text{and}\quad}\)
\(\displaystyle{y}\oplus{x}={\left({y}\cap{x}^{{c}}\right)}\cup{\left({x}\cap{y}^{{c}}\right)}\)
\(\displaystyle={\left({x}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\right)}\)
\(\displaystyle={x}\oplus{y}\)
Associative
For \(\displaystyle{x},{y}{\quad\text{and}\quad}{z}\in{P}{\left({A}\right)}\)
\(\displaystyle{\left({x}\oplus{y}\right)}\oplus{z}={\left({\left({x}\oplus{y}\right)}\cap{z}^{{c}}\right)}\cup{\left({z}{n}{\left({x}\oplus{y}\right)}^{{c}}\right)}\)
\(\displaystyle={\left({z}\cap{\left({\left({x}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\right)}^{{c}}\right)}\cup{\left({\left({\left({x}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\right)}\right)}\cap{z}^{{c}}\right)}\right.}\)
\(\displaystyle={\left({z}\cap{\left({x}^{{c}}\cup{y}\right)}\cap{\left({y}^{{c}}\cup{x}\right)}\right)}\cup{\left({\left({x}\cap{y}^{{c}}\right)}\cap{z}^{{c}}\right)}\cup{\left({\left({y}\cap{x}^{{c}}\right)}\cap{z}^{{c}}\right)}\)
\(\displaystyle={\left({z}\cap{\left({x}^{{c}}\cap{\left({y}^{{c}}\cup{x}\right)}\right)}\cup{\left({\left({y}\cap{y}^{{c}}\right)}\cup{x}\right)}\right)}\cup{\left({\left({\left({x}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\right)}\right)}\cap{z}^{{c}}\right)}\)
\(\displaystyle={\left({z}\cap{\left({\left({x}^{{c}}\cap{y}^{{c}}\right)}\cup{\left({x}^{{c}}\cap{x}\right)}\right)}\cup{\left({\left({y}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}\right)}\right)}\right)}\cup{\left({\left({\left({x}\cap{y}^{{}}\right)}\cap{z}^{{c}}\right)}\cup{\left({\left({y}\cap{x}^{{c}}\right)}\cap{z}^{{c}}\right)}\right)}\)
\(\displaystyle={\left({z}\cap{\left({\left({x}^{{c}}\cap{y}^{{c}}\right)}\cup{\left({x}^{{c}}\cap{x}\right)}\right)}{u}{\left({\left({y}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}\right)}\right)}\right)}\cup{\left({x}\cap{y}^{{c}}\cap{z}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\cap{z}^{{c}}\right)}\)
\(\displaystyle={\left({z}\cap{x}^{{c}}\cap{y}^{{c}}\right)}\cup{\left({z}\cap{y}\cap{x}\right)}\cup{\left({x}\cap{y}^{{c}}\cap{z}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\cap{z}^{{c}}\right)}\)
Similarly, it can be proved that
\(\displaystyle{x}\oplus{\left({y}\oplus{z}\right)}={\left({z}\cap{x}^{{c}}\cap{y}^{{c}}\right)}\cup{\left({z}\cap{y}\cap{x}\right)}\cup{\left({x}\cap{y}^{{c}}\cap{z}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\cap{z}^{{c}}\right)}\)
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