Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}.

Solution:

Given:

A = {1, 2, 4, 5}, B = {2, 3, 5, 6} and C = {4, 5, 6, 7}

We have to verify the following identities:

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

LHS

$(B \cap C)=\{5,6\}$

 

$A \cup(B \cap C)=\{1,2,4,5,6\}$

RHS

$(A \cup B)=\{1,2,3,4,5,6\}$

 

$(A \cup C)=\{1,2,4,5,6,7\}$

$(A \cup B) \cap(A \cup C)=\{1,2,4,5,6\}$

LHS = RHS

$\therefore A \cup(B \cap C)=(A \cup B) \cap(A \cup C)$

(ii) $A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$

LHS

$(B \cup C)=\{2,3,4,5,6,7\}$

 

$A \cap(B \cup C)=\{2,4,5\}$

RHS

$A \cap B=\{2,5\}$

 

$A \cap C=\{4,5\}$

$(A \cap B) \cup(A \cap C)=\{2,4,5\}$

LHS = RHS

$\therefore A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$

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

LHS

$(B-C)=\{2,3\}$

 

$A \cap(B-C)=\{2\}$

RHS

$(A \cap B)=\{2,5\}$

$(A \cap C)=\{4,5\}$

 

$(A \cap B)-(A \cap C)=\{2\}$

LHS = RHS

$\therefore A \cap(B-C)=(A \cap B)-(A \cap C)$

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

LHS

$(B \cup C)=\{2,3,4,5,6,7\}$

 

$A-(B \cup C)=\{1\}$

RHS

$(A-B)=\{1,4\}$

$(A-C)=\{1,2\}$

 

$(A-B) \cap(A-C)=\{1\}$

LHS = RHS

$\therefore A-(B \cup C)=(A-B) \cap(A-C)$

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

LHS

$(B \cap C)=\{5,6\}$

 

$A-(B \cap C)=\{1,2,4\}$

RHS

$(A-B)=\{1,4\}$

$(A-C)=\{1,2\}$

 

$(A-B) \cup(A-C)=\{1,2,4\}$

LHS = RHS

$\therefore A-(B \cap C)=(A-B) \cup(A-C)$

(vi) $A \cap(B \Delta C)=(A \cap B) \Delta(A \cap C)$

LHS

$(B \Delta C)=(B-C) \cup(C-B)$

$(B-C)=\{2,3\}$

 

$(C-B)=\{4,7\}$

$(B-C) \cup(C-B)=\{2,3,4,7\}$

$\Rightarrow(B \Delta C)=\{2,3,4,7\}$

$A \cap(B \Delta C)=\{2,4\}$

RHS

$(A \cap B)=\{2,5\}$

 

$(A \cap C)=\{4,5\}$

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

$(A \cap B)-(A \cap C)=\{2\}$

 

$(A \cap C)-(A \cap B)=\{4\}$

$\{(A \cap B)-(A \cap C)\} \cup\{(A \cap C)-(A \cap B)\}=\{2,4\}$

$\Rightarrow(A \cap B) \Delta(A \cap C)=\{2,4\}$

LHS = RHS

$\therefore A \cap(B \Delta C)=(A \cap B) \Delta(A \cap C)$