Question

For any sets A, B and C, prove that
$\begin{align}
  & [i]\left( A\bigcup B \right)-C=\left( A-C \right)\bigcup \left( B-C \right) \\
 & [ii]\left( A\bigcap B \right)-C=\left( A-C \right)\bigcap \left( B-C \right) \\
\end{align}$

Answer 1

Hint: Prove the R.H.S. of each statement is equal to the L.H.S. of the statement. Use the fact that $A-B=A\bigcap {{B}^{c}}$. Use distributive laws and associative laws of union and intersection to simplify R.H.S. and hence prove L.H.S. is equal to R.H.S.

Complete step-by-step answer:
[i] We have R.H.S. $=\left( A-C \right)\bigcup \left( B-C \right)$
We know that $A-B=A\bigcap {{B}^{c}}$
Hence, we have R.H.S. $=\left( A\bigcap {{C}^{c}} \right)\bigcup \left( B\bigcap {{C}^{c}} \right)$
We know that the intersection of two sets distributes over the union, i.e. $A\bigcap \left( B\bigcup C \right)=\left( A\bigcap B \right)\bigcup \left( A\bigcap C \right)$
Hence, we have
R.H.S. $={{C}^{c}}\bigcap \left( A\bigcup B \right)$
Now, we know that $A-B=A\bigcap {{B}^{c}}$
Hence, we have R.H.S. $=A\bigcup B-C$
Hence, RHS = LHS.

[ii] RHS $=\left( A-C \right)\bigcap \left( B-C \right)$
We know that $A-B=A\bigcap {{B}^{c}}$
Hence, we have R.H.S. $=\left( A\bigcap {{C}^{c}} \right)\bigcap \left( B\bigcap {{C}^{c}} \right)$
We know that the intersection of two sets is associative, i.e. $A\bigcap \left( B\bigcap C \right)=\left( A\bigcap B \right)\bigcap C$
Hence, we have R.H.S. $=A\bigcap \left( {{C}^{C}}\bigcap \left( B\bigcap {{C}^{c}} \right) \right)$
We know that the intersection of two sets is associative, i.e. $A\bigcap \left( B\bigcap C \right)=\left( A\bigcap B \right)\bigcap C$
Hence, we have R.H.S. $=A\bigcap \left( \left( {{C}^{c}}\bigcap {{C}^{c}} \right)\bigcap B \right)$
We know that $A\bigcap A=A$
Hence, we have
R.H.S. $=A\bigcap \left( {{C}^{c}}\bigcap B \right)=A\bigcap \left( B\bigcap {{C}^{c}} \right)$
We know that the intersection of two sets is associative, i.e. $A\bigcap \left( B\bigcap C \right)=\left( A\bigcap B \right)\bigcap C$
Hence, we have
R.H.S. $=\left( A\bigcap B \right)\bigcap {{C}^{c}}$
Now, we know that $A-B=A\bigcap {{B}^{c}}$
Hence, we have
R.H.S. $=A\bigcap B-C$
Hence, LHS = RHS

Note: Verification using Venn diagrams:
[i] Diagram of $A\bigcup B$

Diagram of $A\bigcup B-C$

Diagram of $A-C$

Diagram of $B-C$

Diagram of $\left( A-C \right)\bigcup \left( B-C \right)$

Hence from Venn diagrams, it is verified that $\left( A\bigcup B \right)-C=\left( A-C \right)\bigcup \left( B-C \right)$
Similarly, it can be verified from Venn diagrams that $\left( A\bigcap B \right)-C=\left( A-C \right)\bigcap \left( B-C \right)$