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In some textbook I found a statement like, "Asymmetric is stronger than simply not symmetric".

But as I try to perceive this statement, both appear to be same to me.

For example, parentof is an asymmetric relation. If $A$ is a parentof $B$, $B$ can not be parentof $A$. We arrive at the same conclusion if we call this relation to be not symmetric.

(Liyang Yu. A Developer’s Guide to the Semantic Web. 2nd. Springer, 2014. p 226, last sentence of second paragraph from bottom.)

Big Al
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Masroor
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    Just saying I don't agree this terminology is clear. It's too natural to interchange "a-" and "not" to have distinct meanings for them. – djechlin Jan 27 '16 at 18:24
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    Also "stronger" would imply that "asymmetric" would imply "not symmetric," which is of course not true for the empty relation, based on the answers below. – djechlin Jan 27 '16 at 21:53
  • Never found this terminology. In fact in my mind asymmetric simply means that lacks the property of symmetry, so the relation may have some symmetric pairs and others that aren't. This is also due to the etymology of the word which actually means this. – Bakuriu Jan 28 '16 at 07:52

6 Answers6

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If a relation is symmetric then there is a two way arrow, e.g. if someone is a blood relative of me then I HAVE to be a blood relative of them.

If a relation is not symmetric then there can be one arrow or two, e.g. if I like someone then they may or may not like me, either case could be true.

If a relation is Asymmetric then having one arrow means that there definitely cannot be two, e.g. if someone is my parent then I definitely CANNOT be their parent.

So basically the difference between non-symmetric and asymmetric is that in one we might have two arrows some of the time, but in the other we can NEVER have a second arrow once we have the first.

Big Al
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EHH
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    Isn't this antisymmetric ? – MSalters Jan 28 '16 at 10:50
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    @MSalters No it's not but they are related. Antisymmetric is when there are no two distinct objects with a two way arrow, so two arrows implies the objects are the same. Asymmetry is an example of antisymmetry but it does not cover every possibility so they are distinct relationships. Please see http://math.stackexchange.com/questions/778164/is-an-anti-symmetric-and-asymmetric-relation-the-same-are-irreflexive-and-anti for clarification of this point. – EHH Jan 28 '16 at 13:46
  • Very accessible explanation. Thank you. – M.Y. Babt Apr 03 '16 at 10:53
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It is just the standard negation of quantifiers: Asking that a relation is never true is stronger than the negation of the relation being always true.

A relation being symmetric means that all pairs can be inverted.

The negation of this is that some pairs cannot be inverted.

Asymmetric means that all pairs cannot be inverted.

Big Al
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N. S.
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In asymmetric and antisymmetric relationships no two distinct elements can be joined by arrows in the two directions. Moreover in asymmetric relationships there can be no loops.

Partial orders are antisymmetric. Strict orders are asymmetric.

Both conditions are generally different from the negation of symmetry; the latter means that there is at least one pair of distinct elements with an arrow in one direction but not the other.

Justpassingby
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Let $R$ be some binary relation.

Asymmetric means that: $ \forall x,y \space \space (x,y) \in R \Rightarrow (y,x) \notin R $

Not symmetric means: $ \exists x,y \space \space \space \space (x,y)\in R \space \land (y,x) \notin R $

This means that to prove that a relation is not symmetric, you just have two find two elements for which the property holds. However, proving that a relation is asymmetric means that you have to prove the assymetric property for all possible pairs.

odin19
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One way to see why "not symmetric" isn't the same as "asymmetric" is to look for an example relationship that is neither symmetric nor asymmetric. The relationship $\leq$ is such an example.

That could be the end of my post, but just to clarify:

It is not symmetric because a $\leq$ b does not imply b $\leq$ a. (In other words, you can find values for a and b such that a $\leq$ b and b $\nleq$ a. For example, let a = 1 and b = 2.)

It is not asymmetric because a $\leq$ b does not imply b $\nleq$ a. (In other words, you can find values for a and b such that a $\leq$ b and b $\leq$ a. For example, let a = 1 and b = 1.)

So $\leq$ is not symmetric but isn't asymmetric. And that means "not symmetric" and "asymmetric" are not equivalent.

Big Al
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What the book said is wrong.

Consider such a case: Let <∅, ∅> be an empty set with an empty relation defined on it. Then the empty relation on the empty set must be reflective, irreflective, symmetric, antisymmetric, asymmetric, and transitive simultaneously. Since the relation is both asymmetric and symmetric, it follows that asymmetry cannot deduce non-symmetry.

Indeed, what the book said is right mostly. However, it may fail in some extreme scenarios, like the example presented above.

Pat_Guangtailang
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