Question

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

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. $=(A-C)\bigcup (B-C)$
We know that $A-B=A\bigcap B^c$
Hence, we have R.H.S. $=(A\bigcap C^c)\bigcup (B\bigcap C^c)$
We know that the intersection of two sets distributes over the union, i.e. $A\bigcap (B\bigcup C)=(A\bigcap B)\bigcup (A\bigcap C)$
Hence, we have
R.H.S. $=C^c\bigcap (A\bigcup B)$
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 $=(A-C)\bigcap (B-C)$
We know that $A-B=A\bigcap B^c$
Hence, we have R.H.S. $=(A\bigcap C^c)\bigcap (B\bigcap C^c)$
We know that the intersection of two sets is associative, i.e. $A\bigcap (B\bigcap C)=(A\bigcap B)\bigcap C$
Hence, we have R.H.S. $=A\bigcap (C^C\bigcap (B\bigcap C^c))$
We know that the intersection of two sets is associative, i.e. $A\bigcap (B\bigcap C)=(A\bigcap B)\bigcap C$
Hence, we have R.H.S. $=A\bigcap ((C^c\bigcap C^c)\bigcap B)$
We know that $A\bigcap A=A$
Hence, we have
R.H.S. $=A\bigcap (C^c\bigcap B)=A\bigcap (B\bigcap C^c)$
We know that the intersection of two sets is associative, i.e. $A\bigcap (B\bigcap C)=(A\bigcap B)\bigcap C$
Hence, we have
R.H.S. $=(A\bigcap B)\bigcap C^c$
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 $(A-C)\bigcup (B-C)$

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