Help proving some basic properties of relations

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Homework Help Overview

The discussion revolves around proving properties of relations, specifically focusing on asymmetry, antisymmetry, irreflexivity, and transitivity. Participants are exploring logical implications and relationships between these properties.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish connections between asymmetry and antisymmetry, as well as between irreflexivity and transitivity. They express difficulty in organizing their thoughts and breaking down the proofs.
  • Some participants question the implications of not being irreflexive and how transitivity interacts with the properties being discussed.
  • Others suggest considering contradictions to clarify the relationships between the properties.

Discussion Status

The discussion is ongoing, with participants actively engaging in exploring logical relationships and potential contradictions. Some guidance has been offered regarding the use of proof by contradiction, but no consensus has been reached on the proofs themselves.

Contextual Notes

Participants are grappling with the definitions and implications of the properties of relations, and there is a noted struggle with the logical structure of the proofs. The original poster has requested additional resources for examples of proofs involving these properties.

privyet
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Homework Statement



Prove the following properties of relations:

1) If R is asymmetric then it's antisymmetric.
2) If R is asymmetric then it's irreflexive.
3) If R is irreflexive and transitive then it's asymmetric.

The Attempt at a Solution



1)
If R is asymmetric on a set X, then for all x,y in X: xRy implies \neg(yRx).
If R is antisymetric on X, then for all x,y in X: xRy and yRx implies x = y.

The premise of the antisymmetry relation requires that xRy and yRx but as R is asymmetric we know that \neg(xRy and yRx), therefore given that the premise is false, the conclusion is vacuously true and we can say that if R is asymmetric then it's antisymmetric.

2)If R is asymmetric on a set X, then for all x,y in X: xRy implies \neg(yRx).
If R is irreflexive, then for all x in X, \neg(xRx)

I really don't know how to think about this proof.

3)
If R is irreflexive, then for all x in X, \neg(xRx).
R is transitive if when xRy and yRz for all x,y,z in X, then xRz.

Likewise, I don't know how to begin thinking about this proof.

In all of these problems I'm finding it hard to get my head around the process of doing the proof. If I write the 3rd problem in logic notation I get:
(\forallx \neg(xRx)) \wedge (\forallx,y,z in X xRy \wedge yRz \Rightarrow xRz) \Rightarrow (xRy \Rightarrow \neg(yRx))

How do I break this down and think about it? Any help much appreciated.
 
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With regard to 2), what happens if R is not irreflexive, that is, there exists some x in X such that xRx?

With regard to 3), what does transitivity say about xRy and yRx?
 
Thanks for your reply.

1) I'm not sure, to be honest. One approach I had in mind was to say let x = y, then in the asymmetric relation xRx \Rightarrow \neg(xRx), then using the irreflexive relation I can say that since xRx is the premise it is false and therefore the implication is vacuously true. As I say, I don't feel like I have any idea how to organize my thoughts with regard to this so I have no idea if my statement is correct or not.

2) Transitivity would say that if xRy and yRx then xRx, but this should be false in this case because it is transitive and irreflexive. Am I going in the right direction here?

P.s. If anyone can point to an online resource that gives examples of proofs involving the properties of relations please let me know. I've done a lot of googling but haven't found much that i can apply to these types of questions.
 
privyet said:
Thanks for your reply.
You're welcome.


1) I'm not sure, to be honest. One approach I had in mind was to say let x = y, then in the asymmetric relation xRx \Rightarrow \neg(xRx), then using the irreflexive relation I can say that since xRx is the premise it is false and therefore the implication is vacuously true. As I say, I don't feel like I have any idea how to organize my thoughts with regard to this so I have no idea if my statement is correct or not.
The premise is *not* that R is irreflexive. The premise is that R is asymmetric. Your job is to prove that this means that R is irreflexive. So assume the contrary, that R is not irreflexive. (Hint: Look for a contradiction.)


2) Transitivity would say that if xRy and yRx then xRx, but this should be false in this case because it is transitive and irreflexive. Am I going in the right direction here?
Yes, you are going in the right direction. Transitivity, irreflexivity, and the assumption that there exists some x,y in X such that xRy and yRx leads to a contradiction. One of these things does not go with the other. That assumption has to be false. What does this say about R (Hint #1: It says it's asymmetric, but you have to prove it. Hint #2: Rewrite A→B using 'and' and 'not'.)


You apparently have forgotten about proof by contradiction. It's a powerful tool.
 
Last edited:

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