Answer to Question #88778 in Discrete Mathematics for Shivam Nishad

Question

et U be the set of positive integers 1, 2, 3, ...
etc., A be the set of odd positive integers and
B be the set of even positive integers. Verify
De Morgan's laws.

Answer 1

$$U=\{x\isin\Z^+ \}, A=\{x\isin\Z^+|x\ is\ odd \}, B=\{x\isin\Z^+|x\ is\ even \}$$

$$\overline{A\cup B}=\overline{A}\cap\overline{B}$$

$${A\cup B}=\{x\isin\Z^+|x\ is\ odd \vee x\ is\ even \}=\{x\isin\Z^+ \}=U$$

$$\overline{A\cup B}=\overline{U}=\{ \}$$

$$\overline{A}=\{x\isin\Z^+|x\ is\ not\ odd \}=\{x\isin\Z^+|x\ is\ even \}=B$$

$$\overline{B}=\{x\isin\Z^+|x\ is\ not\ even \}=\{x\isin\Z^+|x\ is\ odd \}=A$$

$$\overline{A}\cap\overline{B}=\{x\isin\Z^+|x\ is\ even\ \wedge \ x\ is\ odd\}=\{ \}$$

Hence

$$\overline{A\cup B}=\overline{A}\cap\overline{B}$$

$$\overline{A\cap B}=\overline{A}\cup\overline{B}$$

$$A\cap B=\{x\isin\Z^+|x\ is\ odd\ \wedge \ x\ is\ even\}=\{ \}$$

$$\overline{A\cap B}=U-\{ \}=U$$

$$\overline{A}=\{x\isin\Z^+|x\ is\ not\ odd \}=\{x\isin\Z^+|x\ is\ even \}=B$$

$$\overline{B}=\{x\isin\Z^+|x\ is\ not\ even \}=\{x\isin\Z^+|x\ is\ odd \}=A$$

$$\overline{A}\cup\overline{B}=B\cup A=\{x\isin\Z^+|x\ is\ even\vee x\ is\ odd \}=\{x\isin\Z^+ \}=U$$

Hence

$$\overline{A\cap B}=\overline{A}\cup\overline{B}$$