Question

Let A, B, and C be sets. Show that (A − B) − C = (A − C) − (B − C).

Short answer

Thus, here we prove it\(\left( {A - B} \right) - C = \left( {A - C} \right) - \left( {B - C} \right)\).

Step-by-step solution

Given

Given in the question A, B and C is a set. Now here we prove it\(\left( {A - B} \right) - C = \left( {A - C} \right) - \left( {B - C} \right)\).

solution

Here we solve this expression\(\left( {A - C} \right) - \left( {B - C} \right) = \left( {A - B} \right) - C\).

Let us solve RHS,

\(\left( {A - C} \right) - \left( {B - C} \right)\)

We know that,

\( \Rightarrow \)\(A - C\)= \(A \cap \mathop C\limits^\_ \)

Here we use this formula to solve our question,

\(\begin{aligned}{c}\left( {A - C} \right) - \left( {B - C} \right) = \left( {A \cap \mathop C\limits^\_ } \right) - \left( {B \cap \mathop C\limits^\_ } \right)\\ = \left( {A \cap \mathop C\limits^\_ } \right) - \left( {\mathop B\limits^\_ \cap C} \right)\\ = \left( {\left( {A \cap \mathop C\limits^\_ } \right) \cap \mathop B\limits^\_ } \right) \cup \left( {\left( {A \cap \mathop C\limits^\_ } \right) \cap C} \right)\\ = \left( {A \cap \left( {\mathop C\limits^\_ \cap \mathop B\limits^\_ } \right)} \right) \cup \left( {A \cap \left( {\mathop C\limits^\_ \cap \mathop C\limits^\_ } \right)} \right)\end{aligned}\)

\(\left( {A - C} \right) - \left( {B - C} \right) = \left( {A \cap \left( {\mathop C\limits^\_ \cap \mathop B\limits^\_ } \right)} \right) \cup \left( {A \cap \phi } \right)\)

We also know that,\(A \cap \phi \)=\(\phi \).

\(\begin{aligned}{c}\left( {A - C} \right) - \left( {B - C} \right) = \left( {A \cap \left( {\mathop C\limits^\_ \cap \mathop B\limits^\_ } \right)} \right) \cup \phi \\ = A \cap \left( {\mathop C\limits^\_ \cap \mathop B\limits^\_ } \right)\\ = \left( {A \cap \mathop B\limits^\_ } \right) \cap \mathop C\limits^\_ \\ = \left( {A - B} \right) - C\end{aligned}\)

Hence ,

RHS = LHS

\(\left( {A - C} \right) - \left( {B - C} \right) = \left( {A - B} \right) - C\).