Question

Let A = {1, 2, 4, 5}, B = {2, 3, 5, 6} and C = {4, 5, 6, 7}. Verify the following identity:
$A-(B\cap C)=(A-B)\cup (A-C)$

Answer 1

Hint: Here, we have to prove that LHS = RHS. For that first consider LHS $A-(B\cap C)$, find $B\cap C$ and then find $A-(B\cap C)$. Similarly for RHS, first find A – B and A – C and then find $(A-B)\cup (A-C)$. The elements will be the same for LHS and RHS.

Complete step-by-step answer:
Here, we are given that A = {1, 2, 4, 5}, B = {2, 3, 5, 6} and C = {4, 5, 6, 7}.

Now, we have to prove the identity:

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

Here, it is sufficient to prove that LHS = RHS.

Now, consider LHS, $A-(B\cap C)$, first we have to find $B\cap C$, the elements which are common for both B and C. Thus,

$\Rightarrow B\cap C=\{5,6\}$

Next, find $A-(B\cap C)$ where we have to find elements which are in A but not in $B\cap C$. Thus,

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

Hence, we got LHS = {1, 2, 4}

Now, consider RHS $(A-B)\cup (A-C)$, first find A – B and A – C .

A – B has elements that are in A but not in B. Thus,

$\Rightarrow A-B=\{1,4\}$

Similarly, we can say that A – C has elements which are in A but not in C.

$\Rightarrow A-C=\{1,2\}$

Next, we have to find $(A-B)\cup (A-C)$, that is to find elements that are either in $A-B$ or $A-C$. Thus,

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

Hence, we have RHS = {1, 2, 4}

Therefore, from equation (1) and equation (2) we can say that LHS = RHS.

Hence, we get the identity $A-(B\cap C)=(A-B)\cup (A-C)$.

Note: Here, we can also prove this with the help of Venn diagram where we have to take three sets A = {1, 2, 4, 5}, B = {2, 3, 5, 6} and C = {4, 5, 6, 7}. Draw these sets as overlapping circles and enter the elements. From the diagram mark $A-(B\cap C)$ and $(A-B)\cup (A-C)$, show that they are equal.