\( ( ( A ^ { \prime } \cup C ^ { \prime } ) \cap B ) ^ { \prime } \quad \cup ( A \cup ( C \cap B ) ^ { \prime } \cup C ) ^ { \prime } \)

To simplify the expression \[ ( ( A ^ { \prime } \cup C ^ { \prime } ) \cap B ) ^ { \prime } \quad \cup ( A \cup ( C \cap B ) ^ { \prime } \cup C ) ^ { \prime }, \] we will use De Morgan's laws and properties of set operations. 1. **First part**: Simplify \( ( ( A ^ { \prime } \cup C ^ { \prime } ) \cap B ) ^ { \prime } \) By De Morgan's law, we have: \[ (X \cap Y)^{\prime} = X^{\prime} \cup Y^{\prime}. \] Thus, \[ ( ( A ^ { \prime } \cup C ^ { \prime } ) \cap B ) ^ { \prime } = ( A ^ { \prime } \cup C ^ { \prime })^{\prime} \cup B^{\prime}. \] Again applying De Morgan's law: \[ ( A ^ { \prime } \cup C ^ { \prime })^{\prime} = A \cap C. \] Therefore, we have: \[ ( ( A ^ { \prime } \cup C ^ { \prime } ) \cap B ) ^ { \prime } = (A \cap C) \cup B^{\prime}. \] 2. **Second part**: Simplify \( ( A \cup ( C \cap B ) ^ { \prime } \cup C ) ^ { \prime } \) Again, using De Morgan's law: \[ (X \cup Y \cup Z)^{\prime} = X^{\prime} \cap Y^{\prime} \cap Z^{\prime}. \] We need to find \( ( C \cap B )^{\prime} \): \[ ( C \cap B )^{\prime} = C^{\prime} \cup B^{\prime}. \] Thus, \[ ( A \cup ( C \cap B )^{\prime} \cup C )^{\prime} = ( A \cup ( C^{\prime} \cup B^{\prime}) \cup C )^{\prime}. \] Applying De Morgan's law: \[ = A^{\prime} \cap (C^{\prime} \cup B^{\prime})^{\prime} \cap C^{\prime}. \] Now, we simplify \( (C^{\prime} \cup B^{\prime})^{\prime} \): \[ (C^{\prime} \cup B^{\prime})^{\prime} = C \cap B. \] Therefore, we have: \[ ( A \cup ( C \cap B )^{\prime} \cup C )^{\prime} = A^{\prime} \cap (C \cap B) \cap C^{\prime}. \] 3. **Combine both parts**: Now we combine the two parts: \[ (A \cap C) \cup B^{\prime} \cup (A^{\prime} \cap (C \cap B) \cap C^{\prime}). \] This expression can be further simplified, but it is already in a relatively simple form. Thus, the final simplified expression is: \[ (A \cap C) \cup B^{\prime} \cup (A^{\prime} \cap C \cap B \cap C^{\prime}). \]