Let `A,B` and `C` be sets such that `phi ne A nn B sube C`. Then which of the following statements is not true?

Options

A. If `(A-C sube)B`, then `AsubeB`
B. `(CuuA)nn(CuuB)=C`
C. If `(A-B)subeC`, then `AsubeC`
D. `BnnC!=phi`

Explanation

To solve the problem, we need to analyze the given sets \( A \), \( B \), and \( C \) under the conditions provided. The conditions state that \( \phi \neq A \cap B \subseteq C \). We need to determine which of the statements regarding these sets is not true.

### Step-by-step Solution:

1. **Understanding the Given Information**:
- We know that \( A \cap B \) is not empty (i.e., \( \phi \neq A \cap B \)).
- We also know that \( A \cap B \) is a subset of \( C \) (i.e., \( A \cap B \subseteq C \)).

2. **Analyzing the Statements**:
- We will evaluate each statement to determine if it holds true under the given conditions.

3. **Statement A**:
- If \( A - C \) is empty, then \( A \subseteq C \).
- Since \( A \cap B \subseteq C \) and \( A \cap B \neq \phi \), it implies that there are elements in \( A \) that are also in \( C \). Thus, \( A \) cannot have elements outside of \( C \) if \( A \cap B \) is not empty. Therefore, this statement is true.

4. **Statement B**:
- \( A \cup C \) is a proper subset of \( C \).
- Since \( A \cap B \subseteq C \), it does not imply that \( A \cup C \) is a proper subset of \( C \). In fact, \( A \cup C \) could equal \( C \) if all elements of \( A \) are already in \( C \). Therefore, this statement is not necessarily true.

5. **Statement C**:
- If \( A - B \subseteq C \), then \( A \subseteq C \).
- Since \( A - B \) represents elements in \( A \) that are not in \( B \), and given that \( A \cap B \subseteq C \), it is possible that all elements of \( A \) are in \( C \). Thus, this statement is true.

6. **Conclusion**:
- The statement that is not true is **Statement B**: \( A \cup C \) is a proper subset of \( C \).

### Final Answer:
The statement that is not true is **Statement B**.