10. Proof: Suppose A, B, and C are any sets. We will show that (A U B) N CCAU(BN C). Suppose x is any element (A U B) N C. By definition of intersection x is in A U Band x is in C. Then by definition...


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Use element argument to prove the statement. Assume that all sets are subsets of a universal set U, Statement: For all sets A, B, and C, A ∩ (B − C) = (A ∩ B) − (A ∩ C). Please do the proof like the way the example proof does it. Thanks.


10. Proof: Suppose A, B, and C are any sets.<br>We will show that (A U B) N CCAU(BN C).<br>Suppose x is any element (A U B) N C.<br>By definition of intersection x is in A U Band x is in C.<br>Then by definition of union x is in A or x is in B, and<br>in both cases x is in C. It follows by definition of<br>union that in case x is in A and x is in C, then x is in<br>AU (B N C) by virtue of being in A. And in case x is<br>in BN C, then x is in A U (B NC) by virtue of being<br>in BN C. Thus in both cases x is in A U (BN C),<br>which proves that every element in (A U B) N C is in<br>AU (B N C).<br>Hence (A U B) nCCAU(BNC) by definition of<br>subset.<br>

Extracted text: 10. Proof: Suppose A, B, and C are any sets. We will show that (A U B) N CCAU(BN C). Suppose x is any element (A U B) N C. By definition of intersection x is in A U Band x is in C. Then by definition of union x is in A or x is in B, and in both cases x is in C. It follows by definition of union that in case x is in A and x is in C, then x is in AU (B N C) by virtue of being in A. And in case x is in BN C, then x is in A U (B NC) by virtue of being in BN C. Thus in both cases x is in A U (BN C), which proves that every element in (A U B) N C is in AU (B N C). Hence (A U B) nCCAU(BNC) by definition of subset.

Jun 05, 2022
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