For example, one of the distributive properties from Theorems 5.18 can be written as follows: For all sets \(X\), \(Y\), and \(Z\) that are subsets of a universal set \(U\), \((X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z).\). Many of these results may be intuitively obvious, but to be complete in the development of set theory, we should prove all of them. Hence, it is verified that set union is associative. Additive Identity: For all \(a \in \mathbb{R}\), \(a + 0 = a = 0 + a\). \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\), Before proving some of these properties, we note that in Section 5.2, we learned that we can prove that two sets are equal by proving that each one is a subset of the other one. So in one case, if \(x \in A\), then \(x \in A \cup B\) and \(x \in A \cup C\). We will prove that \((A \cup B)^c = A^c \cap B^c\) by proving that an element is in \((A \cup B)^c\) if and only if it is in \(A^c \cap B^c\). What does it mean to say that an element \(x\) is in \(B^c\)? On one, shade the region that represents \((A \cup B)^c\), and on the other, shade the region that represents \(A^c \cap B^c\). 1 - 6 directly correspond to identities and implications of propositional logic, and 7 - 11 also follow immediately from them as illustrated below. Use one of De Morgan’s Laws (Theorem 2.8 on page 48) to explain carefully what it means to say that an element \(x\) is not in \(A \cup B\). If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. As part of Theorem 5.20, we proved one of De Morgan’s Laws. If you widh to review them as well as inference rules click here. This proves that \((A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)\). That is, prove that, Let \(A\), \(B\), and \(C\) be subsets of some universal set \(U\). A set is a collection of discrete data items. Commutative Laws: \(a + b = b + a\), for all \(a, b \in \mathbb{R}\). 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Let \(A\), \(B\), and \(C\) be subsets of some universal set \(U\). Then \(x \in A\) or \(x \in B \cap C\). Let \(A\), \(B\), and \(C\) be subsets of some universal set \(U\). We will first prove that \(A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C\). \(A - (B \cup C) = A \cap (B^c \cap C^c)\). \(a \cdot b = b \cdot a\), for all \(a, b \in \mathbb{R}\). \((A \cup B)^c = A^c \cap B^c\), Subsets and Complements \(A \subseteq B\) if and only if \(B^c \subseteq A^c\), We will only prove one of De Morgan’s Laws, namely, the one that was explored in Preview Activity \(\PageIndex{1}\). Progress Check 5.19: Exploring a Distributive Property. The following are the important properties of set operations. \((A \cup B) \cup C = A \cup (B \cup C)\), Distributive Laws \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) If we change the order of writing the elements in a set, it does not make any changes in the set. B n C  =  {1, -2, 3, 4, 5, 6} n {2, 4, 6, 7}, A u (B n C)  =  {0, 1, 2, 3, 4, 6} -----(1), A u B  =  {0, 1, 2, 3, 4} u {1, -2, 3, 4, 5, 6}, A u C  =  {0, 1, 2, 3, 4 } u {2, 4, 6, 7}, (A u B)n(A u C) = {-2, 0, 1, 2, 3, 4, 5, 6}n{0, 1, 2, 3, 4, 6, 7}, (A u B) n (A u C)  =  {0, 1, 2, 3, 4, 6} ---------(1), From the above Venn diagrams (2) and (5), it is clear that. (ii) Let us verify that union is commutative. Two relationships in the next theorem are known as De Morgan’s Laws for sets and are closely related to De Morgan’s Laws for statements. Proof. This proves that \(A \subseteq B\) and hence that (3) implies (1). So let \(y \in (A \cup B) \cap (A \cup C)\). The following are equivalent: To prove that these are equivalent conditions, we will prove that (1) implies (2), that (2) implies (3), and that (3) implies (1). Important Properties of Set Complements. Basic properties of set operations are discussed here. The proofs of the other parts are left as exercises. 2. We can use Venn diagrams to explore the more complicated properties in Theorem 5.18, such as the associative and distributive laws. What does it mean to say that an element \(x\) is in \(A^c\)? This means that \(x \in (A \cup B) \cap (A \cup C)\). Compare your response in Part (3) to your response in Part (5). So in both cases, we see that \(x \in (A \cup B) \cap (A \cup C)\), and this proves that \(A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C\). \(\emptyset ^c = U\) and \(U^c = \emptyset\), De Morgan's Laws \((A \cap B)^c = A^c \cup B^c\) We will prove that \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\) by proving that each set is a subset of the other set. Let \(x \in A \cap B\). and, hence, that \((A \cup B)^c = A^c \cap B^c\). For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. However, we also know that if \(S\) and \(T\) are both subsets of a universal set \(U\), then.