Problem 59 - Discrete Mathematics with Applications

Question

Simplify each set expression. $$\left(A-B^{\prime}\right)-\left(B-A^{\prime}\right)$$

Step-by-Step Solution

Answer

The simplified set expression is: \(\boxed{\left( A \cap B^{\prime} \right) \cap B^{\prime} \cup \left( A \cap B^{\prime} \right)}\)

Step 1: Understand the given expression

The expression is given as: \[ \left(A - B^{\prime} \right) - \left( B - A^{\prime} \right) \] Here, A and B are sets, and A' and B' are the complements of A and B, respectively. The main operation in the expression is subtraction, which can be rewritten as a union using the definition of set difference: \(A - B = A \cap B^{\prime}\).

Step 2: Rewrite the expression using the definition of set difference

Using the definition of set difference, the given expression can be rewritten as: \[ \left( A \cap B^{\prime} \right) \cap \left( B - A^{\prime} \right)^{\prime} \] Now, we need to simplify this expression further by addressing the right part of the expression \((B - A^{\prime})^{\prime}\).

Step 3: Simplify the complement of the difference

Using De Morgan's Law, we can rewrite \((B - A^{\prime})^{\prime}\) as: \[ \left( B \cap A^{\prime} \right)^{\prime} = B^{\prime} \cup \left( A^{\prime} \right)^{\prime} \] Since, \(A^{\prime\prime} = A\) for any set A, we have: \[ \left( B \cap A^{\prime} \right)^{\prime} = B^{\prime} \cup A \] Now, let's substitute this simplified term in the main expression.

Step 4: Substitute simplified term and rewrite the expression

Substituting the simplified term in the main expression: \[ \left( A \cap B^{\prime} \right) \cap \left( B^{\prime} \cup A \right) \] Now, we can use the Distributive Law of Sets to further simplify the expression.

Step 5: Apply Distributive Law and simplify the expression

Applying Distributive Law of Sets: \[ \left( A \cap B^{\prime} \right) \cap B^{\prime} \cup \left( A \cap B^{\prime} \right) \cap A \] As \((A \cap B^{\prime}) \cap A = A \cap B^{\prime}\), because A is the common element, the final simplification gives: \[ \left( A \cap B^{\prime} \right) \cap B^{\prime} \cup \left( A \cap B^{\prime} \right) \] The simplified set expression is: \[ \boxed{\left( A \cap B^{\prime} \right) \cap B^{\prime} \cup \left( A \cap B^{\prime} \right)} \]

Key Concepts

Set Difference

The idea of a set difference is fundamental in set theory. When we talk about "subtracting" one set from another, what we're really doing is finding which elements belong to the first set but not the second.
To put it mathematically, if you have two sets, say set \( A \) and set \( B \), the set difference \( A - B \) (or sometimes \( A \setminus B \)) consists of all elements that are in \( A \) but not in \( B \):

• Represented as \( A - B = \{ x : x \in A \text{ and } x otin B \} \).

This concept helps separate out unique properties of sets and can simplify complex expressions when used correctly. Mastery of this basic operation is fundamental in understanding more complicated set operations.

De Morgan's Laws

De Morgan's Laws are two essential principles that connect set operations with logical operators. They provide a method to simplify the complement of complex set operations.
Here's how they work, broken down into easily digestible parts:

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

Understanding these transformations can make seemingly convoluted set identities much more manageable. These laws are especially powerful when dealing with large set expressions or proving set identities.

Set Complement

The complement of a set is the collection of elements not in the given set. It's as if you're highlighting what is "outside" the set boundaries relative to a 'universal set,' which includes all possible elements in the context.
Mathematically, if you have a set \( A \), its complement \( A' \) is defined as:

• \( A' = \{ x : x otin A \} \)

One key property of complements is that the complement of a complement returns the original set: \((A')' = A \). This cycle ensures that no matter how many times an element is included or excluded, these operations will balance each other out when applied in sequence.
Understanding complements is crucial when simplifying set expressions, as they provide insights into addressing what parts of the set universe are not covered by the set itself.

Distributive Law of Sets

The Distributive Law in set theory is akin to distributing factors in arithmetic but applied to operations with set intersections and unions. It provides a powerful tool for simplifying complex set expressions.
It has two forms, working with the building blocks of set operations:

• The intersection distributes over the union: \( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \).
• The union distributes over the intersection: \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \).

These principles are useful for breaking down expressions into manageable pieces, allowing for deeper analysis and simplification. Mastery of these laws empowers problem-solvers to elegantly and effortlessly maneuver through the complexities of set algebra.