Question

For any two sets A and B, prove the following :

(i) $A\cap (A^{\prime }\cup B)=A\cap B$

(ii) A - (A - B) = $A\cap B.$

(iii) $A\cap (A\cup B{)}^{\prime }=\varphi$

(iv) A - B =$A\Delta (A\cup B)$.

Answer 1

(i) LHS = $A\cap (A\cup B)$

= $(A\cap A^{\prime })\cup (A\cap B)$

[$\because \cap$ distributes over (i)]

= $\varphi \cup (A\cap B)$ [$\because A\cap A^{\prime }=\varphi$]

= $A\cap B$ [$\because \varphi \cup x=x$ for any set x]

= RHS

$\therefore$ LHS = RHS Proved.

(ii) For any sets A and B we have By De-morgan's laws

$(A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime },(A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }$

Also,

LHS = A- (A - B)

= $A\cap (A-B{)}^{\prime }$

=$A\cap (A-B^{\prime }{)}^{\prime }$

= $A\cap [A^{\prime }\cup (B^{\prime })$]' [By De-morgan's laws]

= $A\cap (A^{\prime }\cup B)$ [$\because (B^{\prime }{)}^{\prime }=B$]

= $(A\cap A^{\prime })\cup (A\cap B)$

= $\varphi \cup (A\cap B)$ [$\because A\cap A^{\prime }=\varphi$]

= $A\cap B$ [$\because \varphi \cup x=x,$ for any set x]

= RHS

$\therefore$ LHS = RHS Proved.

(iii) LHS = $A\cap (A\cup B^{\prime })$

= $A\cap (A^{\prime }\cap B^{\prime })$ [By De-morgan's law]

= $(A\cap A^{\prime })\cap B^{\prime }$ [By associative law]

= $\varphi \cap B^{\prime }$ [$\because A\cap A^{\prime }=\varphi$]

= $\varphi$

= RHS

$\therefore$ LHS = RHS Proved.

(iv) RHS = $A\Delta (A\cap B)$

= {$A-(A\cap B)$} \(\cup (A \cap B - A)

[$\because E\Delta F=(F-F)\cup (F-E)$]

= {$A\cap (A\cap B{)}^{\prime }$] $\cup (A\cap B\cap A^{\prime }$ } [$\because E-F=E\cap F^{\prime }$]

= { $A\cap (A^{\prime }\cap B^{\prime })$] $\cup (A\cap A^{\prime }\cap B$}

[By De-morgan's law and associative law]

= $(A\cap A^{\prime })\cup (A\cap B^{\prime })\cup (\varphi \cap B)$

[$\because \cap$ distributes over $\cup andA\cap A^{\prime }=\varphi$]

= $\varphi \cup (A\cap B^{\prime })\cup \varphi$ [$\because \varphi \cap B=\varphi$]

= $A\cap B^{\prime }$ [$\because \varphi \cup x=x$ for any set x]

= A - B [$\because A\cap B^{\prime }=A-B$]

= LHS

$\therefore$ LHS = RHS Proved.