/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Beweisen Sie das Distributivgese... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 20

Beweisen Sie das Distributivgesetz: $$ M_{1} \cup\left(M_{2} \cap M_{3}\right)=\left(M_{1} \cup M_{2}\right) \cap\left(M_{1} \cup M_{3}\right) $$

Short Answer

Expert verified
Question: Prove that the Distributive Law holds for sets by showing that \(M_{1} \cup\left(M_{2} \cap M_{3}\right)=\left(M_{1} \cup M_{2}\right) \cap\left(M_{1} \cup M_{3}\right)\).

Step by step solution

01

Prove subset 1 (from left to right)

To prove that \(M_1 \cup (M_2 \cap M_3) \subseteq (M_1 \cup M_2) \cap (M_1 \cup M_3)\), we must show that for any element \(x \in M_1 \cup (M_2 \cap M_3)\), \(x\) also belongs to \((M_1 \cup M_2) \cap (M_1 \cup M_3)\). Assume that \(x \in M_1 \cup (M_2 \cap M_3)\). Then, by definition of the union, \(x \in M_1\) or \(x \in (M_2 \cap M_3)\). We need to show that \(x \in (M_1 \cup M_2) \cap (M_1 \cup M_3)\). Case 1: \(x \in M_1\). In this case, by definition of the union, \(x\) is in both \(M_1 \cup M_2\) and \(M_1 \cup M_3\). Therefore, \(x\) is in their intersection \((M_1 \cup M_2) \cap (M_1 \cup M_3)\). Case 2: \(x \in (M_2 \cap M_3)\). In this case, \(x\) is in both \(M_2\) and \(M_3\). Hence, \(x\) is in the union \(M_1 \cup M_2\) and also in the union \(M_1 \cup M_3\). Therefore, \(x\) is in their intersection \((M_1 \cup M_2) \cap (M_1 \cup M_3)\). In both cases, we found that \(x \in (M_1 \cup M_2) \cap (M_1 \cup M_3)\). Thus, the first subset is proven.
02

Prove subset 2 (from right to left)

To prove that \((M_1 \cup M_2) \cap (M_1 \cup M_3) \subseteq M_1 \cup (M_2 \cap M_3)\), we must show that for any element \(x \in (M_1 \cup M_2) \cap (M_1 \cup M_3)\), \(x\) also belongs to \(M_1 \cup (M_2 \cap M_3)\). Assume that \(x \in (M_1 \cup M_2) \cap (M_1 \cup M_3)\). Then, by definition of the intersection, \(x \in M_1 \cup M_2\) and \(x \in M_1 \cup M_3\). We need to show that \(x \in M_1 \cup (M_2 \cap M_3)\). Case 1: \(x \in M_1\). In this case, by definition of the union, \(x\) is in \(M_1 \cup (M_2 \cap M_3)\). Case 2: \(x \notin M_1\). In this case, \(x\) must be in both \(M_2\) and \(M_3\) since it is in the unions \((M_1 \cup M_2)\) and \((M_1 \cup M_3)\). Therefore, \(x\) is in the intersection \(M_2 \cap M_3\). Hence, \(x \in M_1 \cup (M_2 \cap M_3)\). In both cases, we found that \(x \in M_1 \cup (M_2 \cap M_3)\). Thus, the second subset is proven.
03

Conclusion

Since we have proven both \(M_1 \cup (M_2 \cap M_3) \subseteq (M_1 \cup M_2) \cap (M_1 \cup M_3)\) and \((M_1 \cup M_2) \cap (M_1 \cup M_3) \subseteq M_1 \cup (M_2 \cap M_3)\), we conclude that the sets are equal, which means that the Distributive Law holds for these sets: $$ M_{1} \cup\left(M_{2} \cap M_{3}\right)=\left(M_{1} \cup M_{2}\right) \cap\left(M_{1} \cup M_{3}\right) $$

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Most popular questions from this chapter

Die Menge \(A_{4}\) hat vier Elemente, die Mengen \(B_{3}\), \(B_{4}\) und \(B_{5}\) haben entsprechend drei, vier und f眉nf Elemente. 脺berlegen Sie jeweils, ob es Abbildungen $$ \begin{aligned} &f_{43}: A_{4} \rightarrow B_{3} \\ &f_{44}: A_{4} \rightarrow B_{4} \\ &f_{45}: A_{4} \rightarrow B_{5} \end{aligned} $$ geben kann, die (a) injektiv, aber nicht surjektiv, (b) surjektiv, aber nicht injektiv, (c) bijektiv sind.

Es seien \(M_{1}\) und \(M_{2}\) Teilmengen von \(X\). Beweisen Sie die einfachste Form der Regeln von de Morgan, wobei wir \(C_{X}\) als Bezeichnung f眉r die Komplementbildung bez眉glich \(X\) verwenden: $$ \begin{aligned} &C_{X}\left(M_{1} \cap M_{2}\right)=C_{X}\left(M_{1}\right) \cup C_{X}\left(M_{2}\right) \\ &C_{X}\left(M_{1} \cup M_{2}\right)=C_{X}\left(M_{1}\right) \cap C_{X}\left(M_{2}\right) \end{aligned} $$ Stellen Sie diesen Sachverhalt mittels Venn-Diagrammen dar.

Beweisen Sie die Abtrennregel (modus ponens): $$ (A \wedge(A \Rightarrow B)) \Rightarrow B $$

Beweisen Sie die 脛quivalenzen: $$ \begin{aligned} (A \vee B) & \Leftrightarrow \neg(\neg A \wedge \neg B) \\ (A \wedge B) & \Leftrightarrow \neg(\neg A \vee \neg B) \\ (A \Rightarrow B) & \Leftrightarrow((\neg A) \vee B) \end{aligned} $$

Wie viele unterschiedliche bin盲re, also zwei Aussagen verkn眉pfende Junktoren gibt es?
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