For all sets [tex]$A, B$[/tex], and [tex]$C$[/tex], construct an algebraic proof for the given statement. Cite a property from Theorem 6.22 for every step.

30. [tex]$(A \cap B) \cup C=(A \cup C) \cap(B \cup C)$[/tex]
31. For all sets [tex]$A$[/tex] and [tex]$B, A \cup(B-A)=A \cup B$[/tex].

We proved the first identity by applying the distributive property, showing that ( A ∩ B ) ∪ C = ( A ∪ C ) ∩ ( B ∪ C ) . The second identity was established by rewriting B − A as B ∩ A c and then applying the distributive property and the complement property to demonstrate that A ∪ ( B − A ) = A ∪ B . Both identities effectively illustrate key properties of set operations.
;

Answered by Anonymous | 2025-07-04

Prove the first identity: ( A c a pB ) c u pC = ( A c u pC ) c a p ( B c u pC ) by applying the distributive property.
Prove the second identity: A c u p ( B − A ) = A c u pB by rewriting B − A as B c a p A c .
Apply the distributive property to get ( A c u pB ) c a p ( A c u p A c ) .
Use the complement property and identity property to simplify to A c u pB , thus proving the identity.

Explanation

Problem Analysis We are given two set identities to prove using algebraic proofs and properties from Theorem 6.22. The identities are:

For all sets A , B , and C , ( A c a pB ) c u pC = ( A c u pC ) c a p ( B c u pC )

For all sets A and B , A c u p ( B − A ) = A c u pB .

Starting with RHS Let's prove the first identity: ( A c a pB ) c u pC = ( A c u pC ) c a p ( B c u pC ) .


We will start with the right-hand side (RHS) and show that it is equal to the left-hand side (LHS).
( A c u pC ) c a p ( B c u pC )

Applying Distributive Property Apply the distributive property: ( A c u pC ) c a p ( B c u pC ) = ( A c a pB ) c u pC . This is the distributive property of intersection over union.

Conclusion for Identity 1 Thus, we have shown that ( A c u pC ) c a p ( B c u pC ) = ( A c a pB ) c u pC , which is the left-hand side.

Rewriting the Left-Hand Side Now, let's prove the second identity: A c u p ( B − A ) = A c u pB .


We know that B − A = B c a p A c , where A c is the complement of A . So, we can rewrite the left-hand side as:
A c u p ( B − A ) = A c u p ( B c a p A c )

Applying Distributive Property Apply the distributive property: A c u p ( B c a p A c ) = ( A c u pB ) c a p ( A c u p A c )

Using Complement Property Use the complement property: A c u p A c = U , where U is the universal set. So, we have:


( A c u pB ) c a p ( A c u p A c ) = ( A c u pB ) c a p U

Using Identity Property Use the identity property: ( A c u pB ) c a p U = A c u pB . Thus, we have shown that A c u p ( B − A ) = A c u pB .

Final Answer Therefore, we have proven both set identities using algebraic proofs and properties from Theorem 6.22.


Examples
Set theory is used in computer science, particularly in database management and software engineering. For example, when designing a database, set operations can help in querying and manipulating data. In software engineering, sets can represent collections of objects or states, and set operations can be used to manage these collections efficiently. Understanding set identities helps in optimizing database queries and simplifying software logic, leading to more efficient and reliable systems. For instance, proving that ( A c a pB ) c u pC = ( A c u pC ) c a p ( B c u pC ) allows us to rewrite a complex query in a database to a simpler, more efficient form without changing the result.

Answered by GinnyAnswer | 2025-07-04