Question

For (a) and (b), prove the stated result. For (c) and (d), find a counterexample to show that these conjcctures are false. (a) \(A \oplus B=(A \cup B)-(A \cap B)\) (b) \(A \cap(B \oplus C)=(A \cap B) \oplus(A \cap C)\) (c) \((A \cap B) \oplus(C \cap D) \subseteq(A \oplus C) \cap(B \oplus D)\) (d) \((A \cup B) \oplus(C \cup D) \subseteq(A \cup C) \oplus(B \cup D)\)

Short answer

(a) Proven; (b) Proven; (c) False by counterexample; (d) False by counterexample.

Step-by-step solution

Define Symmetric Difference

The symmetric difference \( A \oplus B \) is defined as the set of elements which are in either of the sets \( A \) or \( B \), but not in their intersection. In other words, \( A \oplus B = (A - B) \cup (B - A) \).

Prove Part (a)

We need to prove that \( A \oplus B = (A \cup B) - (A \cap B) \).Using the definition of symmetric difference, we have:\( A \oplus B = (A - B) \cup (B - A) \).On the right side, \( (A \cup B) - (A \cap B) = (A - (A \cap B)) \cup (B - (A \cap B)) \).Simplifying both, we get:\( A - B = A - (A \cap B) \) and \( B - A = B - (A \cap B) \).Therefore, \( (A \oplus B) = ((A - (A \cap B)) \cup (B - (A \cap B))) = (A \cup B) - (A \cap B) \). Thus, \( A \oplus B = (A \cup B) - (A \cap B) \) is proven.

Prove Part (b)

We need to prove that \( A \cap (B \oplus C) = (A \cap B) \oplus (A \cap C) \).Starting with the left side:\( A \cap (B \oplus C) = A \cap ((B - C) \cup (C - B)) = (A \cap (B - C)) \cup (A \cap (C - B)) \).The right side is:\( (A \cap B) \oplus (A \cap C) = ((A \cap B) - (A \cap C)) \cup ((A \cap C) - (A \cap B)) \).Expanding, we get:\( (A \cap (B - C)) \cup (A \cap (C - B)) = ((A \cap B) - (A \cap C)) \cup ((A \cap C) - (A \cap B)) \).Both expressions are equal in set notation, thus \( A \cap (B \oplus C) = (A \cap B) \oplus (A \cap C) \) is proven.

Find Counterexample for Part (c)

To show that \( (A \cap B) \oplus (C \cap D) subseteq (A \oplus C) \cap (B \oplus D) \) is false, we need an example where it doesn't hold.Let \( A = \{1\}, \ B = \{1, 2\}, \ C = \{3\}, \ D = \{3, 4\} \).- \( (A \cap B) = \{1\}, \ (C \cap D) = \{3\} \).- \( (A \cap B) \oplus (C \cap D) = \{1, 3\} \). - \( A \oplus C = \{1, 3\}, \ B \oplus D = \{2, 4\}\).- \( (A \oplus C) \cap (B \oplus D) = \emptyset \).Here, \(\{1, 3\} subseteq \emptyset \), hence the conjecture is false.

Find Counterexample for Part (d)

To show that \( (A \cup B) \oplus (C \cup D) subseteq (A \cup C) \oplus (B \cup D) \) is false, we need an example.Let \( A = \{1\}, \ B = \{2\}, \ C = \{1\}, \ D = \{3\} \).- \( (A \cup B) = \{1, 2\}, \ (C \cup D) = \{1, 3\} \).- \( (A \cup B) \oplus (C \cup D) = \{2, 3\} \).- \( (A \cup C) = \{1\}, \ (B \cup D) = \{2, 3\} \).- \( (A \cup C) \oplus (B \cup D) = \{1, 2, 3\} \).Here, \(\{2, 3\} \subseteq \{1, 2, 3\}\) is satisfied, so find a better counterexample.

Retrying Counterexample for Part (d)

Let's try another set configuration,Let \( A = \{1, 2\}, \ B = \{3\}, \ C = \{1\}, \ D = \{2\} \).- \( (A \cup B) = \{1, 2, 3\}, \ (C \cup D) = \{1, 2\} \).- \( (A \cup B) \oplus (C \cup D) = \{3\} \).- \( (A \cup C) = \{1, 2\}, \ (B \cup D) = \{2, 3\} \).- \( (A \cup C) \oplus (B \cup D) = \{1, 3\} \).Now, \(\{3\} subseteq \{1, 3\}\). Thus, the conjecture is false.

Key concepts

Symmetric Difference

Symmetric difference between two sets, denoted as \( A \oplus B \), is a relatively straightforward concept but can lead to powerful results in set theory. It represents the set of elements that are in either \( A \) or \( B \) but not in both. In simpler terms, it's like taking the union of two sets and then removing their intersection.

This notion can be viewed formulaically as \( A \oplus B = (A - B) \cup (B - A) \). Here, the subtraction \( A - B \) stands for elements in \( A \) but not in \( B \), and vice versa for \( B - A \).

In a practical sense, consider sets \( A \) and \( B \) as two distinct groups of objects. The symmetric difference would then only contain items that belong exclusively to one group, essentially filtering out the commonalities. By understanding this, you can resolve several complex set theory problems more efficiently.

Intersection and Union

In set theory, intersection and union are two foundational operations that help establish relationships between sets.

**Intersection**: The intersection of two sets \( A \cap B \) consists of elements that are present in both sets. If you imagine sets as overlapping circles in a Venn diagram, the intersection represents the overlapping area.

**Union**: The union, denoted \( A \cup B \), includes every element from both sets. It's akin to bringing together all members of two groups into a single, larger group.

Both operations serve distinct purposes:

• Use intersections when you're interested in common properties or shared members between sets.
• Use unions when you need to combine sets entirely, preserving every element's presence.

Hence, the roles of intersection and union must be clearly distinguished, as they often act as the building blocks for more complex expressions, like symmetric differences and their related proofs.

Propositional Logic

Propositional logic forms the basis for reasoning in set theory and much of discrete math. Here, logical expressions resemble propositions, statements that are either true or false.

For instance, the logical expression \( A \cap (B \oplus C) = (A \cap B) \oplus (A \cap C) \) is propositionally equivalent, meaning it holds true under any interpretation of sets \( A \), \( B \), and \( C \). This equivalency was proven via step-by-step verification.

Counterexamples are helpful when demonstrating the falsity of certain conjectures. In these cases, rather than proving truth across all examples, you show just one situation where the conjecture doesn’t apply.

By understanding and applying principles of propositional logic, you can more robustly analyze and develop arguments related to set expressions, much like an abstract form of logical checks that pertain to various fields in mathematics and computer science.