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Prove that for any three sets A, B, and C we have

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

Please help. I do not know where to start.

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    The LHS is a set. The RHS is a set. How do you prove that two sets are equal? – 2017-02-13
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    You might draw a Venn diagram, and work from there. – 2017-02-13
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    Were you provided with some definitions? Maybe some other theorems, and proofs thereof? For example, do you have some other theorem or proof that shows that two sets are equal? Because then you can try and follow the same strategy that was used there for this proof. – 2017-02-13
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    I'm assuming you are familiar with the operations of union, intersection, and complementation. Are you familiar with DeMorgan's law http://www.geocities.ws/nate623a/traffcomp/demorgans-law.jpg – 2017-02-13
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    If you are familiar with [absolute complements](https://en.wikipedia.org/wiki/Complement_(set_theory)#Absolute_complement) write the LHS as $A \,\cap\, \overline{B \cap \overline{C}}$ then apply [De Morgan's laws](https://en.wikipedia.org/wiki/De_Morgan's_laws). – 2017-02-13

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Start with "suppose $x \in A - (B-C)$". Then try to show that $x$ must be in $(A-B)\cup (A\cap C)$. Then do the same thing in the other direction. You will then have proved the LHS and RHS are subsets of each other, meaning they must be equal.

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first you can check the equation for special cases like $B=C$ or $A=(B-C)$

one way to deal with simple problems of this type is to use the indicator functions of the sets. for simplicity let $\chi_A$ be the indicator function for the set $A$, so $\chi_A(x)=1$ if $x\in A$ and $0$ otherwise.

you require to know how the boolean operations on sets are represented as algebraic operations. thus: $$ \chi_{A\cap B} = \chi_A \chi_B \\ \chi_{A\cup B}=\chi_A +\chi_B - \chi_A\chi_B \\ \chi_{A-B} = \chi_A(1-\chi_B) $$ using these relations you can see that $$ \begin{align} \chi_{A-(B-C)} & = \chi_A(1-\chi_{B-C}) \\ & = \chi_A(1-\chi_B(1-\chi_C)) \\ & = \chi_A - \chi_A\chi_B +\chi_A\chi_B\chi_C \\ \end{align} $$ on the other side: $$ \begin{align} \chi_{(A-B)\cup(A\cap C)} & = \chi_{(A-B)}+\chi_{(A\cap C)} - \chi_{(A-B)}\chi_{(A\cap C)} \\ & = \chi_{(A-B)}+\chi_{(A\cap C)} - \chi_A(1-\chi_B)\chi_A\chi_C \end{align} $$ you should be able to complete the answer now, as long as you remember that an indicator function is idempotent, i.e. for any set $A$, $\chi_A^2=\chi_A$