Explanation Given information: P is a set of posets. Proof: A poset ( X , 1 ≺ ) 1 is isomorphic to another poset ( X , 2 ≺ ) 2 , if there is one to one correspondence f : X → 1 X 2 , such that a ≺ b 1 ⇔ f ( a ) ≺ f 2 ( b ) Isomorphism is reflexive; every poset is isomorphic to itself under the identity function. If ( x , 1 ≺ ) 1 is isomorphic to another poset ( x , 2 ≺ ) 2 and f is the isomorphism, then it is one-to-one correspondence, Such that: ( a ≺ b 1 ) ⇔ f ( a ) ≺ f 2 ( b ) . Hence f − 1 : X 2 → X 1 is also a one-to-one correspondence. a ≺ b 2 ⇔ f − 1 ( b ) Isomorphism is symmetric

3rd Edition

ISBN: 9781284056242

Author: David J. Hunter

Publisher: Jones & Bartlett Learning

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Chapter 2.5, Problem 31E

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To prove:

The relation ≅ is an equivalence relation on P, where P is a set of posets.

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Chapter 2.5, Problem 30E

Chapter 2.6, Problem 1E

Chapter 2 Solutions

Essentials of Discrete Mathematics

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - Prob. 14E Ch. 2.2 - Prob. 15E Ch. 2.2 - Prob. 16E Ch. 2.2 - Prob. 17E Ch. 2.2 - Prob. 18E Ch. 2.2 - Prob. 19E Ch. 2.2 - Prob. 20E Ch. 2.2 - Prob. 21E Ch. 2.2 - Prob. 22E Ch. 2.2 - Prob. 23E Ch. 2.2 - Prob. 24E Ch. 2.2 - Prob. 25E Ch. 2.2 - Prob. 26E Ch. 2.2 - Prob. 27E Ch. 2.2 - Prob. 28E Ch. 2.2 - Prob. 29E Ch. 2.2 - Prob. 30E Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.3 - Prob. 1E Ch. 2.3 - Prob. 2E Ch. 2.3 - Prob. 3E Ch. 2.3 - Prob. 4E Ch. 2.3 - Prob. 5E Ch. 2.3 - Prob. 6E Ch. 2.3 - Prob. 7E Ch. 2.3 - Prob. 8E Ch. 2.3 - Prob. 9E Ch. 2.3 - Prob. 10E Ch. 2.3 - Prob. 11E Ch. 2.3 - Prob. 12E Ch. 2.3 - Prob. 13E Ch. 2.3 - Prob. 14E Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 16E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - Prob. 21E Ch. 2.3 - Prob. 22E Ch. 2.3 - Prob. 23E Ch. 2.3 - Prob. 24E Ch. 2.3 - Prob. 25E Ch. 2.3 - Prob. 26E Ch. 2.3 - Prob. 27E Ch. 2.3 - Prob. 28E Ch. 2.3 - Prob. 29E Ch. 2.3 - Prob. 30E Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.4 - Prob. 1E Ch. 2.4 - Prob. 2E Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - Prob. 5E Ch. 2.4 - Prob. 6E Ch. 2.4 - Prob. 7E Ch. 2.4 - Prob. 8E Ch. 2.4 - Prob. 9E Ch. 2.4 - Prob. 10E Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - Prob. 14E Ch. 2.4 - Prob. 15E Ch. 2.4 - Prob. 16E Ch. 2.4 - Prob. 17E Ch. 2.4 - Prob. 18E Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - Prob. 21E Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - Prob. 24E Ch. 2.4 - Prob. 25E Ch. 2.4 - Prob. 26E Ch. 2.4 - Prob. 27E Ch. 2.4 - Prob. 28E Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Prob. 32E Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.4 - Prob. 37E Ch. 2.4 - Prob. 38E Ch. 2.4 - Prob. 39E Ch. 2.4 - Prob. 40E Ch. 2.5 - Prob. 1E Ch. 2.5 - Prob. 2E Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - Prob. 5E Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Prob. 20E Ch. 2.5 - Prob. 21E Ch. 2.5 - Prob. 22E Ch. 2.5 - Prob. 23E Ch. 2.5 - Prob. 24E Ch. 2.5 - Prob. 25E Ch. 2.5 - Prob. 26E Ch. 2.5 - Prob. 27E Ch. 2.5 - Prob. 28E Ch. 2.5 - Prob. 29E Ch. 2.5 - Prob. 30E Ch. 2.5 - Prob. 31E Ch. 2.6 - Prob. 1E Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - Prob. 7E Ch. 2.6 - Prob. 8E Ch. 2.6 - Prob. 9E Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Prob. 23E Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Prob. 29E Ch. 2.6 - Prob. 30E

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To prove: The relation ≅ is an equivalence relation on P , where P is a set of posets.

Solution Summary: The author explains that the relation cong is an equivalence relation on P, where P is a set of posets. Isomorphism is reflexive,

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To prove:

The relation ≅ is an equivalence relation on P, where P is a set of posets.

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Chapter 2 Solutions

To prove: The relation ≅ is an equivalence relation on P , where P is a set of posets.

Solution Summary: The author explains that the relation cong is an equivalence relation on P, where P is a set of posets. Isomorphism is reflexive,

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To prove: The relation ≅ is an equivalence relation on P, where P is a set of posets.

Want to see the full answer?

Chapter 2 Solutions

To prove:

The relation ≅ is an equivalence relation on P, where P is a set of posets.