Question

For any two sets A and B, prove that

(i) $(A\cup B)-B=A-B$

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

(iii) $A-(A-B)=A\cap B$

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

(v) $(A-B)\cup (A\cap B)=A$

Answer 1

(i) $(A\cup B)-B=A-B$

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

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

= $A\cap B^{\prime }$

= A- B = RHS

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

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

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

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

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

= A - B = RHS

(iii)$A-(A-B)=A\cap B$

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

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

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

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

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

= $A\cap B$ = RHS

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

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

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

= $A\cup B=RHS$ [$\because A\cap U=A$]

(v) $(A-B)\cup (A\cap B)=A$

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

= [$(A-B)\cup \left[\right(A-B)\cup B$]

= $A\cap (A\cup B)=A=RHS$