Show that $A\subseteq B, a\in A , a\notin (B-C)\Rightarrow a\in C$.

Question

Question : Show that $$A\subseteq B, a\in A , a\notin (B-C)\Rightarrow a\in C$$

My Try : Let $a\notin C$ , we know $a\in A$ and $A\subseteq B$ so $a\in B$ . And we can see that $a \in B-C$ This is a contradiction . This solution is not as if clear. How one can solve it . This solution is true.Right?

Answer 1

Your solution is correct. A typical more formal proof might look like this:

1. $a \in A \quad{\text{[Given]}}$
2. $A \subset B \quad{\text{[Given]}}$
3. $a \notin (B- C) \quad{\text{[Given]}}$
4. By Contradiction: suppose $a \notin C \quad{\text{[Contradiction Hypothesis]}}$
5. $a \in B \quad{\text{[Reason 1, 2]}}$
6. $a \in (B - C) \quad{\text{[Reason 4, 5, Definition of set-difference.]}}$
7. Contradiction $\quad{\text{[Reason 3, 6 disagree]}}$
8. Hence the contradiction hypothesis 4 is false, so $a \in C$.