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This problem concerns the following six sets: $$\begin{array}{c}A=\\{0,2,4,6\\} \quad B=(1,3,5) \quad C=\\{0,1,2,3,4,5,6,7\\} \\\D=\emptyset \quad E=\mathbb{N} \quad F=\\{10,2,4,6\\} \mid\end{array}$$ (a) What sets are subsets of \(A\) ? (b) What sets are subsets of \(B\) ? (c) What sets are subsets of \(C\) ? (d) What sets are subsets of \(D\) ? (c) What sets are subsets of \(E\) ? (f) What sets are subsets of \(F\) ?

Short Answer

Expert verified
(a) \(D\); (b) \(D\); (c) \(A, B, D, F\); (d) \(D\); (e) \(A, B, C, D, F\); (f) \(A, D\).

Step by step solution

01

Understanding Subsets

A subset is a set in which every element of the subset is also an element of the larger set. We need to compare each given set against the other sets to determine if they qualify as a subset.
02

Subsets of Set A

Set \( A = \{0,2,4,6\} \). To find subsets of \( A \), check which sets have all their elements present in \( A \). Upon evaluation, \( D = \emptyset \), containing no elements, is a subset by definition of any set. Set \( F = \{10,2,4,6\} \), but since 10 is not in \( A \), it is not a subset. No other sets fit the criteria.
03

Subsets of Set B

Set \( B = (1,3,5) \). Evaluate other sets. Again, \( D \) is a subset of any set because it is empty. No other sets are subsets of \( B \) as they do not contain the exact elements \(1, 3, 5\).
04

Subsets of Set C

Set \( C = \{0,1,2,3,4,5,6,7\} \). All the sets that either have no elements or have elements that only appear in \( C \) are its subsets. Hence, \( A, B, D, \) and \( F \) are subsets of \( C \).
05

Subsets of Set D

Set \( D = \emptyset \). By definition, the empty set is a subset of every set, including itself. So, \( D \) is a subset of \( D \).
06

Subsets of Set E

Set \( E = \mathbb{N} \), the set of natural numbers. Any set containing natural numbers could be considered a subset of \( E \). In this list, \( A, B, C, D \), and \( F \) fit the criteria as they consist of non-negative integers.
07

Subsets of Set F

Set \( F = \{10,2,4,6\} \). Apply the subset rule: \( D \) with no elements can always fit. Set \( A \) is also a subset as it only contains \( 2, 4, 6 \).
08

Final Step: Compile Results

Based on these evaluations, we determine that: - Subsets of \( A \): \( D \).- Subsets of \( B \): \( D \).- Subsets of \( C \): \( A, B, D, F \).- Subsets of \( D \): \( D \).- Subsets of \( E \): \( A, B, C, D, F \).- Subsets of \( F \): \( A, D \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Empty Set
The concept of the empty set, denoted by - \( \emptyset \), represents a set with no elements whatsoever. It's like having an empty box—there's nothing inside it. This special set is unique because it is considered a subset of every possible set. - Whether the set is full of elements or also empty, the empty set fits comfortably as a subset. This is because there are no elements in it to contradict the contents of another set. - By definition, a set can only be a subset if every element in the set is also present in another set. Since there are zero elements in an empty set, it vacuously satisfies this condition for any set.
- This explains why, during our subset evaluations: * The empty set \( \emptyset \) is listed as a subset of set \( A \), set \( B \), set \( C \), set \( D \), set \( E \), and set \( F \). * It represents universality in the world of subsets.
Working with Natural Numbers
The set of natural numbers, commonly denoted by \( \mathbb{N} \), refers to the set of all positive integers that we use for counting. - - This includes numbers like 0, 1, 2, 3, and so on (although in some contexts it begins with 1). It goes on infinitely, representing every non-negative integer. - Natural numbers form the foundation of basic arithmetic and are crucial in understanding subset relations, especially when dealing with the infinite nature of \( \mathbb{N} \).
- When evaluating subsets within different sets, recognizing that \( \mathbb{N} \) encompasses a wide range allows us to confidently identify subsets based on their elements' membership in \( \mathbb{N} \). * This means anything that is a non-negative integer in the provided sets is a subset of \( E = \mathbb{N} \). * Sets like \( A = \{0,2,4,6\} \) or \( B = (1,3,5) \), which contain elements from natural numbers, are subsets of \( \mathbb{N} \). * Similarly, the empty set \( D \) and the set \( F = \{10,2,4,6\} \) are subsets of natural numbers because they either contain non-negative integers or no elements at all.
Understanding the Subset Rule
The subset rule is a simple yet crucial concept in set theory. For one set to be a subset of another, every element in the first set must be found within the second set. - Think of it like reading off a list of items that should match another list perfectly, though the first list can have fewer items. - - This concept is symbolically represented as \( A \subseteq B \), meaning set \( A \) is a subset of set \( B \).
- Applying this rule allows us to determine subset relationships by checking if each subset candidate's elements are included in the potential larger set: * For example, if we have the set \( A = \{0,2,4,6\} \), the empty set \( D \) is a subset because it has no conflicting elements. * A set like \( F = \{10,2,4,6\} \) isn’t a subset because 10 isn’t included in \( A \). * Understanding and applying this rule helps in analyzing and identifying subsets accurately in any given situation.

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Most popular questions from this chapter

Write pseudocode for a program that, given a formula \(\phi,\) finds (i) a logically equivalent formula \(\phi^{\prime}\) in CNF and (ii) a logically equivalent formula \(\phi^{\prime \prime}\) in DNF. The algorithm should be recursive (similar to an induction on formulas) and should not involve the construction of truth tables. Prove the algorithm works. This gives an alternate proof of the theorem that every formula is equivalent to a formula in CNF.

Let \(A, B,\) and \(C\) be sets. (a) Prove that if \(A \subset B\) and \(B \subseteq C\), then \(A \subset C\). (b) Prove that if \(A \subseteq B\) and \(B \subset C,\) then \(A \subset C\). (c) Prove that if \(A \subseteq B\) and \(A \not \subseteq C,\) then \(B \nsubseteq C\).

Write the truth tables for the following formulas, Use the truth table to determine whether any of these formulas is a tautology. (a) \(((p \rightarrow q) \wedge(q \rightarrow r)) \rightarrow(p \leftrightarrow r)\) (b) \(((p \rightarrow q) \wedge(q \rightarrow r)) \rightarrow(p \rightarrow r)\) (c) \(((p \rightarrow q) \rightarrow r) \rightarrow(p \rightarrow(q \rightarrow r))\) (d) \((p \rightarrow(r \vee q)) \rightarrow((p \rightarrow r) \vee(p \rightarrow q))\) (e) \((p \rightarrow(r \wedge q)) \rightarrow((p \rightarrow r) \vee(p \rightarrow q))\) (f) \(((p \rightarrow q) \rightarrow q) \rightarrow p\)

Given an array Values with \(n\) elements Values[0], Values[1],.... Values \([n-1 \mid\) each containing a real number, the following algorithm finds the sum of all the positive values in Values. Write an invariant for the loop. rollingSum \(=0\) for \(i=\phi, 2, \ldots, n-1\) if Values \([t]>0\) rolling Sum \(=\) rollingSum \(+\) Values \([i]\) Output rollingSum.

The length of a clause is the number of literals in the clause. The length of a CNF formula is the sum of the length of its clauses. The number of excess literals in a CNF formula is the length of the formula minus the number of clauses in the formula. (a) Show that if an unsatisfiable set \(S\) of clauses contains only clauses of length 0 and 1 , it has a resolution refutation. (Hint: Prove the following: If \(S\) contains a clause of length 0 , it has [trivially] a resolution refutation. If, for some proposition letter \(p, S\) contains both \(p\) and \(\neg p,\) then \(S\) has a resolution refutation. Otherwise, \(S\) is satisfiable.) (b) Show that if a set \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k} \vee \lambda_{k+1}+\cup S(k \geq 1)\right.\) of clauses is un- satisfiable, so are \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \lambda_{k}\right\\} \cup S\) and \(\left\\{\lambda_{k+1}\right\\} \cup S\). (Hint: For the first half, prove that if an interpretation \(I\) satisfies \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k}\right\\} \cup S,\) it also satisfies \(\left.\left\\{\lambda_{1} \vee \lambda_{2} \vee \cdots \vee \lambda_{k} \vee \lambda_{k+1}\right\\} \cup S_{.}\right)\) (c) Show that for \(k \geq 1,\) the number of excess literals in \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \cdots \vee \lambda_{k}\right\\} \cup S\) and the number of excess literals in \(\left\\{\lambda_{k+1}\right\\} \cup S\) are both less than the number of excess literals in \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k} \vee \lambda_{k+1}\right\\} \cup S\). (d) A resolution derivation of a clause \(r_{k}\) from a set \(S\) of clauses is a sequence \(r_{0}, r_{1}, r_{2}, \ldots, r_{k}\) of clauses where each \(r_{l}\) is either an element of \(S\) or a resolvant of two previous \(r\) 's. (Thus, resolution refutation of \(S\) is just a resolution derivation of \(F\) from \(S .\) ) Show that if there is a resolution derivation of \(\lambda\) from \(S\) and a resolution refutation of \(S \cup\\{\lambda\\},\) then there is a resolution refutation of \(S .\) (e) Prove that if there is a resolution refutation \(\rho\) of \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k}\right\\} \cup S,\) then either (i) there is a resolution refutation of \(\left.\mid \lambda_{1} \vee \lambda_{2} \vee \cdots \vee \lambda_{k} \vee \lambda_{k+1}\right\\} \cup S\) or (ii) there is a resolution derivation of \(\lambda_{k+1}\) from \(\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k} \vee \lambda_{k+1} \cup S\). (Hint: Prove this by induction on the length \(\rho\). You will have to add \(\lambda_{k+1}\) as a disjunct to some of the clauses in \(\rho\). It is not true in general that if \(S \models \lambda\), then there is a resolution derivation of \(\lambda\) from \(S .\) ) (f) Prove that resolution refutation is complete.

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