In 30-40 words, construct an algebraic proof for the given statement. Give a property from Theorem 6.22 for every step.

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

(1) If all sets [tex]$A$[/tex] and [tex]$B$[/tex], [tex]$A \cup(B-A)=A \cup B$[/tex].

We proved that ( A ∩ B ) ∪ C = ( A ∪ C ) ∩ ( B ∪ C ) by establishing both subset inclusions. First, we showed that if an element is in the left side, it must be on the right side, and vice versa. Thus, the two expressions represent the same set.
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Answered by Anonymous | 2025-07-04

Assume an arbitrary element x is in ( A c a pB ) c u pC , then show x must be in ( A c u pC ) c a p ( B c u pC ) .
Assume an arbitrary element x is in ( A c u pC ) c a p ( B c u pC ) , then show x must be in ( A c a pB ) c u pC .
By showing membership in both directions, prove the set identity.
Conclude that ( A c a pB ) c u pC = ( A c u pC ) c a p ( B c u pC ) .

Explanation

Problem Analysis We are tasked with proving the set identity ( A c a pB ) c u pC = ( A c u pC ) c a p ( B c u pC ) using an algebraic proof. This means we need to show that any element in ( A c a pB ) c u pC is also in ( A c u pC ) c a p ( B c u pC ) and vice versa. We will use properties from Theorem 6.22 to justify each step.

Setting up the Proof Let x be an arbitrary element. We want to show that x s u b se t e q ( A c a pB ) c u pC if and only if x s u b se t e q ( A c u pC ) c a p ( B c u pC ) . This will prove the equality of the two sets.

Part 1: Showing ( A c a pB ) c u pC s u b se t e q ( A c u pC ) c a p ( B c u pC ) First, let's show that if x s u b se t e q ( A c a pB ) c u pC , then x s u b se t e q ( A c u pC ) c a p ( B c u pC ) .


Assume x s u b se t e q ( A c a pB ) c u pC . This means x s u b se t e q ( A c a pB ) or x s u b se t e qC (by the definition of union).
Case 1: x s u b se t e q ( A c a pB ) . This means x s u b se t e q A or x s u b se t e qB (by the definition of intersection).
If x s u b se t e q A , then x s u b se t e q A c u pC . Also, x may or may not be in B c u pC .
If x s u b se t e qB , then x s u b se t e qB c u pC . Also, x may or may not be in A c u pC .
However, since x s u b se t e q ( A c a pB ) c u pC , we have x s u b se t e q ( A c a pB ) or x s u b se t e qC . If x s u b se t e qC , we proceed to Case 2.
So, if x s u b se t e q A then x s u b se t e q A c u pC . If x s u b se t e qB then x s u b se t e qB c u pC .
Case 2: x s u b se t e qC . Then x s u b se t e q A c u pC and x s u b se t e qB c u pC . Therefore, x s u b se t e q ( A c u pC ) c a p ( B c u pC ) .
Combining both cases, we can say that if x s u b se t e q ( A c a pB ) c u pC , then x s u b se t e q ( A c u pC ) c a p ( B c u pC ) .

Part 2: Showing ( A c u pC ) c a p ( B c u pC ) s u b se t e q ( A c a pB ) c u pC ) Now, let's show that if x s u b se t e q ( A c u pC ) c a p ( B c u pC ) , then x s u b se t e q ( A c a pB ) c u pC .

Assume x s u b se t e q ( A c u pC ) c a p ( B c u pC ) . This means x s u b se t e q ( A c u pC ) and x s u b se t e q ( B c u pC ) (by the definition of intersection).
So, x s u b se t e q A c u pC means x s u b se t e q A and x s u b se t e qC (by the definition of union). And x s u b se t e qB c u pC means x s u b se t e qB and x s u b se t e qC (by the definition of union).
Since x s u b se t e q A and x s u b se t e qB , we have x s u b se t e q ( A c a pB ) (by the definition of intersection). Also, we know x s u b se t e qC .
Therefore, x s u b se t e q ( A c a pB ) or x s u b se t e qC , which means x s u b se t e q ( A c a pB ) c u pC (by the definition of union).
Thus, if x s u b se t e q ( A c u pC ) c a p ( B c u pC ) , then x s u b se t e q ( A c a pB ) c u pC .

Conclusion Since we have shown that x s u b se t e q ( A c a pB ) c u pC if and only if x s u b se t e q ( A c u pC ) c a p ( B c u pC ) , we can conclude that ( A c a pB ) c u pC = ( A c u pC ) c a p ( B c u pC ) . This completes the algebraic proof.

Examples
Understanding set identities like ( A c a pB ) c u pC = ( A c u pC ) c a p ( B c u pC ) is crucial in database management. Imagine A represents customers who bought product X, B represents customers who bought product Y, and C represents customers who have a premium membership. The left side, ( A c a pB ) c u pC , represents customers who bought both X and Y, or have a premium membership. The right side, ( A c u pC ) c a p ( B c u pC ) , represents customers who either bought X or have a premium membership, AND either bought Y or have a premium membership. This identity ensures that both descriptions target the same group of customers, which is vital for marketing campaigns and data analysis.

Answered by GinnyAnswer | 2025-07-04