Question

The set \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap\left(\mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right)^{\prime} \cap \mathrm{C}^{\prime}\) equals (a) \(\mathrm{B} \cap \mathrm{C}^{\prime}\) (b) \(\mathrm{B} \cup \mathrm{C}^{\prime}\) (c) \(\mathrm{A} \cap \mathrm{C}\) (d) \(\mathrm{A} \cup \mathrm{C}\)

Short answer

The short answer based on the given step-by-step solution is: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap\left(\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}'\right)' \cap \mathrm{C}' = \mathrm{B} \cap \mathrm{C}'\)

Step-by-step solution

Apply De Morgan's law

To simplify the expression, apply De Morgan's law to \( (\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}')' \). This law states that: \( (A \cap B)’ = A’ \cup B’ \). Therefore, we have: \((\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}')' = (\mathrm{A}')\cup(\mathrm{B}'')\cup(\mathrm{C}'') \) As we know \( P''=P \), the expression becomes: \((\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}')' = (\mathrm{A}')\cup \mathrm{B} \cup \mathrm{C} \)

Distribute the intersection

Now, utilize the distributive property of intersections to simplify further. The distributive property states that: \((A ∩ (B ∪ C)) = (A ∩ B) ∪ (A ∩ C)\). Thus, the given expression is: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}'\) To distribute the intersection: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}' = (\mathrm{A}\cap (\mathrm{A}'\cup \mathrm{B}\cup \mathrm{C})\cap \mathrm{C}' )\cup (\mathrm{B}\cap (\mathrm{A}'\cup \mathrm{B}\cup \mathrm{C})\cap \mathrm{C}' )\cup (\mathrm{C}\cap (\mathrm{A}'\cup \mathrm{B}\cup \mathrm{C})\cap \mathrm{C}' )\).

Simplify with set properties

We will now use several set properties to simplify the expression. The properties are: 1. \(A \cap A’ = ∅\) 2. \(A \cap ∅ = ∅\) 3. \(A \cap B = B \cap A\) 4. \(A \cap (B ∪ C) = (A ∩ B) ∪ (A ∩ C)\) Apply these properties to the expression: \((\mathrm{A} \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}') = ∅ \) \((\mathrm{B} \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}') = (\mathrm{B} \cap \mathrm{A}') \cup (\mathrm{B}\cap\mathrm{B})\cup (\mathrm{B} \cap \mathrm{C})\cap \mathrm{C'\) ) Following set properties 1 and 2: \((\mathrm{B} \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}') = (\mathrm{B} \cap \mathrm{A}') \cup ∅ \cup (\mathrm{B} \cap \mathrm{C}\cap \mathrm{C'})\) This simplifies to: \((\mathrm{B} \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}') = (\mathrm{B} \cap \mathrm{A}') \) \((\mathrm{C}\cap (\mathrm{A}'\cup \mathrm{B}\cup \mathrm{C})\cap \mathrm{C}' ) = ∅\) Thus, we have: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}' = (\mathrm{B} \cap \mathrm{A}')\)

Identify the correct option

Comparing the simplified expression with the given options, we see that: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}' = (\mathrm{B} \cap \mathrm{A}')\) Which is equal to option (a): \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap\left(\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}'\right)' \cap \mathrm{C}' = \mathrm{B} \cap \mathrm{C}'\)

Key concepts

De Morgan's Laws

De Morgan's Laws are fundamental in set theory and logic. They provide a way to simplify and manipulate complex expressions involving the intersection and union of sets. These laws are expressed as follows:

• The complement of the intersection of two sets is equal to the union of their complements: \((A \cap B)' = A' \cup B'\).
• The complement of the union of two sets is equal to the intersection of their complements: \((A \cup B)' = A' \cap B'\).

De Morgan's Laws help in transforming expressions to make calculations simpler. For example, in the original exercise, we applied De Morgan's Law to convert \((\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}')'\) into \((\mathrm{A}') \cup \mathrm{B} \cup \mathrm{C}\). This simplification is crucial for further computations as it allows us to work with easier expressions. It's a powerful tool for simplifying complex boolean expressions.

Intersection of Sets

The intersection of sets is another key concept in set theory. When we talk about intersection, denoted by a \((\cap)\) symbol, we mean the set containing all elements that are common to the sets being intersected. For example, \(A \cap B\) contains all elements that are present in both set \(A\) and set \(B\).

• This requires an understanding that only the 'overlapping' parts of the sets are included in the result.
• For disjoint sets, the intersection would result in an empty set, denoted by \(\emptyset\).

In the context of the exercise, understanding intersection helps to identify where sets have mutual elements, which is important in simplifying expressions further. By simplifying intersections, we achieve a more manageable form of the equation for further analysis.

Distributive Property in Sets

The distributive property in sets is similar to its role in arithmetic, allowing for expressions to be rearranged and simplified. In set theory, the distributive property states:

• \((A \cap (B \cup C)) = (A \cap B) \cup (A \cap C)\)
• Conversely, \((A \cup (B \cap C)) = (A \cup B) \cap (A \cup C)\)

Using the distributive property helps to break down the complex expressions into parts that are simpler to understand. In our solution, by applying this property to intersected unions and vice versa, we rearrange parts of the expression for efficient computation and simplification. It’s especially useful when dealing with combined set operations, as seen in the example where distributing intersection over union led us to isolate more intuitive terms and eventually find simpler intersections.

Set Operations Simplification

Simplification of set operations involves using various rules and properties to reduce an expression to its simplest form. This process typically involves:

• Applying laws like De Morgan's and properties such as associative and distributive.
• Eliminating unnecessary terms (like intersections with an empty set).
• Recognizing certain identities, such as \(A \cap \emptyset = \emptyset\) and \(A \cup \emptyset = A\).

In the exercise given, set simplification allowed us to break down a complex initial expression to something as simple as \((\mathrm{B} \cap \mathrm{A}')\), which can be compared directly to possible answers. This process is essential in problem-solving within set theory, helping to ensure that expressions are not only simplified but logically equivalent to their original form.