# set operations proofs

\] The empty set, $$\emptyset$$, is the set with no elements: ... Logicians sometimes describe ordinary mathematical proofs as informal, in contrast to the formal proofs in natural deduction. \mathbf{R} = \mathbf{Q} \cup \overline{\mathbf{Q}}\,.\], Written $$A\cap B$$ and defined But from the definition of set difference, we see that though they can be proven also using some of these properties (after those properties are proven, needless to say). Theorem: For any sets, $$\overline{A\cup B}= \overline{A} \cap \overline{B}$$. 13. . Then the set of students taking classes this semester is When writing informal proofs, the focus is on readability. Hence . Proof for 11: Let x be an arbitrary element in the universe. The set $$\overline{B}$$ is the set of all values not in $$B$$. Theorem: $$A-(B\cup C)= (A-B)\cap(A-C)$$. but also for others. &= \{x\mid x\in A \wedge x\in \overline{B}\} \\ when we're working with real numbers, probably $$U=\mathbf{R}$$. Back to Schedule We can use the set identities to prove other facts about sets. Hence . For example, (b) can be proven as follows: A &= \{x\mid x \in (\overline{A}\cap\overline{B}) )\} \\ If , then and . 1 - 6 directly correspond (See section 2.2 example 10 for that. x A A The standard notation for irrational numbers should now make a lot of sense: with universal set $$\mathbf{R}$$, the irrationals ($$\overline{\mathbf{Q}}$$) are the complement of the rationals ($$\mathbf{Q}$$). x\in S \wedge x\notin{S}\,. x Back to Table of Contents. A $A\cup B\cup C \cup D\,,\\A\cap B\cap C \cap D\,.$. A to identities &= \{x\mid x\in (A \cap \overline{B})\} \\ If we need to do union/intersection of a lot of things, there is a notation like summation that is used occasionally. Theorem: For any sets, $$|A\cap B|\le|A|$$ and $$|A\cap B|\le|B|$$. B B, by 7 6. and that of -------     Identity Laws 5. is also in This name is used since the basic method is to choose an arbitrary element from one set and “chase it” until you prove it must be in another set. \overline{A\cup B} \] Alternative proof For example: Those identities should convince you that order of unions and intersections don't matter (in the same way as addition, multiplication, conjunction, and disjunction: they're all commutative operations). First by 15   by the commutativity of 4. ( cf. )   by the definition of set union. Proof for 4: and Proof for 12: (a) ? Proof for 8: (a) If then . The primary purpose of this section is to have in one place many of the properties of set operations that we may use in later proofs. &= \overline{A}\cap\overline{B}\,.\quad{}∎ ) by the definition of ( B - A ) . Since , since . A Since (use "addition" rule), and between follows. The properties 1 6 , and 11 Thus, A−B = A∩Bc. By deﬁnition of set diﬀerence, x ∈ A− B. by the definition of . ( B - A ) 2. 12. if and only if and Proof for 13: Since , . &= \{x\mid \neg(x \in A)\wedge \neg(x\in B )\} \\ Properties of Set Operation Subjects to be Learned . 9. Theorem For any sets A and B, A∩B ⊆ A. B Often not explicitly defined, but implicit based on the problem we're looking at. ( B Since we're doing the same manipulations, we ended up with the same tables. A-B -------     Domination Laws $\sum_{i=1}^{n} i^2\,.$. and vice versa. if and only if We are going to prove this by showing that every element that is in by 1. Let the sets $$S_1,S_2,\ldots ,S_n$$ be the students in each course. For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. = A by 3, For example, suppose there are $$n$$ courses being offered at ZJU this semester. Like the domain for quantifiers, it's the set of all possible values we're working with. Hence . Next -- Recursive Definition Notice the similarity between the corresponding set and logical operators: $$\vee,\cup$$ and $$\wedge,\cap$$ and $$\overline{\mbox{S}},\neg$$. B -------     Distributive Laws We have used the choose-an-element method to prove Propositions 5.7, 5.11, and 5.14. \overline{\mathbf{Q}} = \mathbf{R}-\mathbf{Q} \,.\]. Look familiar? Then there must be an element $$x$$ with $$x\in(A-B)$$, but $$x\notin A$$. by the definition \begin{align*} 1. Then by the definition of the operators, Hence A satisfies the conditions for the complement of . The “more formal” version has more steps and leaves out the intuitive reason (that might help you actually remember why). Alternative proof: This proof might give a hint why the equivalences and set identities tables are so similiar. Less Formal Proof: The set $$A-B$$ is the values from $$A$$ with any values from $$B$$ removed. Note here the correspondence of Here is an example. = ( A (See example 10 for an example of that too.). B ) Here the only if part is going to be proven. &= A\cap \overline{B}\cap A\cap \overline{C} \\ Be careful with the other operations. of propositional logic, and 7 - 11 also follow immediately from them as illustrated below. &= \{x\mid x\notin (A\cup B)\} \\ With similar proofs, we could prove these things: When doing set operations we often need to define a. ( cf. ) \[\bigcup_{i=1}^{n} S_i\,. There is no logical version of set difference, or set version of exclusive or (at least as far as we have defined). x \end{align*}. Since A Additional properties: Then there is an element x that is in , i.e. For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: A. The if part can be proven similarly. Hence . If and , then , Also since , . the commutativity of Proof: Suppose for contradiction that there is an element $$x\in S\cap\overline{S}$$. x Could have also given a less formal proof. Proof: By definition of the set operations, A These can also be proven using 8, 14, and 15. equalities involving set operations intersection of sets subset relations proofs of equalities proofs of subset relations Contents . Proof for 9: Let x be an arbitrary element in the universe. by the definition of . Set. A. x\in S \wedge x\in\overline{S} \\ \[A\cap B = \{x \mid x\in A\wedge x\in B\}\,\\ Theorem For any sets A and B, B ⊆ A∪ B. Proof… by the distribution Also . A Hence . B . Proof for 6: By the definition of the equality of sets, we need to prove that Let x be an arbitrary element in the universe. Here are some basic subset proofs about set operations. Hence does not hold. Since , . • Applying this to S we get: • x (x S x S) which is trivially True • End of proof Note on equivalence: • Two sets are equal if each is a subset of the other set. Copyright © 2013, Greg Baker. if and only if

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