Why is a symmetric relation defined by $\forall{x}\forall{y}(xRy \implies yRx)$ and not $\forall{x}\forall{y}(xRy \iff yRx)$? (I have only found a couple of sources that defines it with a biconditional)

For example, according to Wolfram:

A relation $R$ on a set $S$ is symmetric provided that for every $x$ and $y$ in $S$ we have $xRy \iff yRx$.

But the majority of books defines it the other way. And I think I agree with the second definition.

Because if we use the first definition with "$\implies$", we know the truth table of the implication in particular $P \implies Q$ is true when $P$ is false and $Q$ is true. That means in the context of symmetric relation that $(x,y) \notin R \implies (y,x) \in R$ is true.

And the example $A = \{1,2,3,4\}$ with relation $R = \{(2,1),(3,1),(4,1)\}$ satisfies the definition because $(x,y) \notin R \implies (y,x) \in R$ is true.

And for me it's weird that this case is considered symmetric. Or maybe I have a profound confusion with the concept. I would like that you guys help me clarify. *Sorry for my grammar I'm not a native english speaker.

  • $\begingroup$ It's for ALL x, y. In your example $(2,1)\in R \not \implies (1,2)\in R$. But given $A=\{1,2,3\}$ and $R=\{(1,2),(2,1)\}$ we have $(1,3)\in R \implies (3,1)\in R$ etc. $\endgroup$ – fleablood Mar 23 at 15:34
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    $\begingroup$ "That means in the context of symmetric relation that (x,y)∉ R ⟹ (y,x)∈ R is true" Only if $x,y$ is actually in $R$. $(2,1)\not \in R \implies (1,2)\in R$ is a true statement. But $(3,2)\not \in R \implies (2,3)\in R$ is a false statement. $\endgroup$ – fleablood Mar 23 at 15:44
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    $\begingroup$ The definitions are equivalent. $\endgroup$ – PyRulez Mar 23 at 18:56

For all $x$ and all $y$ make the if and only if unnecessary (albeit perfectly acceptable).

1) $(x,y) \in R \implies (y,x) \in R$ for ALL $x,y \in A$

And the statement 2) $(x,y) \in R \iff (y,x) \in R$ are equivalent statements.

If 1) is true and $(x,y) \not \in R$ then although $(x,y)\in R\implies (y,x)\in R$ or $F \implies (y,x)\in R$ is true, it does not tell us any thing about whether or not $(y,x) \in R$. However $(y,x) \in R \implies (x,y) \in Y$ tells us that $(y,x) \not \in R$. Because $(y,x) \in R \implies (x,y) \in R$ means $(y,x) \in R \implies F$. An the only thing that implies a false statement is a false statement. So we must have $(y,x) \not \in R$.

So in your example you have $(1,2)\in R\implies (2,1)\in R$ is true but you don't have $(2,1) \in R \implies (1,2) \in R$ as true.

So it isn't symmetric.


Another way to look at it:

If $A = \{1,2,3\}$

Then we will have 9 statments.

By 1) the nine statements are:

$(1,1)\in R\implies (1,1) \in R$

$(1,2) \in R \implies (2,1) \in R$

$(1,3) \in R \implies (3,1) \in R$

$(2,1) \in R \implies (1,2) \in R$

... etc... all nine are needed.

With 2) we also have nine statements:

$(1,1)\in R\iff (1,1) \in R$

$(1,2) \in R \iff (2,1) \in R$

$(1,3) \in R \iff (3,1) \in R$

$(2,1) \in R \iff (1,2) \in R$


$(1,2) \in R \iff (2,1) \in R$ and $(2,1) \in R \iff (1,2)\in R$ is redundant.

So aesthetically, using definition 2) is .... inefficient.

  • $\begingroup$ Got it. Thank you, fleablood. $\endgroup$ – Rodrigo Sango Mar 23 at 16:16
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    $\begingroup$ "all nine are needed" ... except the three that are tautologies, of course. But it would be more work and less clear to exclude them than to tolerate them. $\endgroup$ – Henning Makholm Mar 23 at 18:57
  • $\begingroup$ Well, none of them are needed if you state it as all $(x,y)$ but that includes, for each $(x,y)$, $(y,x)$ as well so... well, why say more than you need to. $\endgroup$ – fleablood Mar 24 at 2:14

If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $A\times A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(\forall x\in A)(\forall y\in A):x\mathrel Ry\implies y\mathrel Rx.\tag1$$And this is equivalent to$$(\forall x\in A)(\forall y\in A):x\mathrel Ry\iff y\mathrel Rx.\tag2$$So, why do we choose $(1)$ instead of $(2)$ in general? Because, in general (although not in this case) it is easier to verify the assertion $A\implies B$ than $A\iff B$. And (again, in general), when we choose between two distinct but equivalent definitions, we usually choose the one which is easier to verify that it holds.


If $xRy \implies yRx$ for all $x$ and all $y$, then we can choose $x := \tilde{y}$ and $y := \tilde{x}$ and get $\tilde{y}R\tilde{x} \implies \tilde{x}R\tilde{y}$ or, equivalently, $yRx \implies xRy$, which is $\impliedby$.

In conclusion, since the implication should hold for all $x,y$, the equivalence already holds.


Let it be that $R$ is a symmetric relation.

This according to the first mentioned definition:$$\forall x\forall y[xRy\implies yRx]\tag1$$

Now let it be that $aRb$.

Then we are allowed to conclude that $bRa$.

On the other hand if $bRa$ then also we are conclude that $aRb$.

So apparantly we have:$$aRb\iff bRa$$

Proved is now that for a symmetric relation $R$ (based on definition $(1)$) we have:$$\forall x\forall y[xRy\iff yRx]\tag2$$

So $(2)$ is a necessary condition for $(1)$.

Next to that it is obvious that $(2)$ is also a sufficient condition for $(1)$ so actually the statements are equivalent.

Both can be used as definition then, but in cases like that it is good custom to go for the one with less implications. This for instance decreases the chance on redundant work if we try to prove that a relation is indeed symmetric.


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