Why is a symmetric relation defined: $forall xforall y( xRyimplies yRx)$ and not $forall xforall y (xRyiff yRx)$?How to prove relation is asymmetric if it is both anti-symmetric and irreflexiveIs an Anti-Symmetric Relation also Reflexive?Graph, Relation $xRy Leftrightarrow$ There is a path between $x$ and $y$ - symmetryDefine symmetric relation R on set SProve that if $R$ is a symmetric, transitive relation on $A$ and the domain of $R$ is $A$, then $R$ is reflexive on $A$.If R is symmetric, must $S=forall xin X exists yin Y (xRy) $ be symmetric?Given set A, is the relation A x A always anti symmetric?Definition of symmetric relationName for relation property: If $xRy$ and $xRz$ and $x not =y$, then $yRz$.Can a relation be transitive when it is symmetric but not reflexive?

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Why is a symmetric relation defined: $forall xforall y( xRyimplies yRx)$ and not $forall xforall y (xRyiff yRx)$?


How to prove relation is asymmetric if it is both anti-symmetric and irreflexiveIs an Anti-Symmetric Relation also Reflexive?Graph, Relation $xRy Leftrightarrow$ There is a path between $x$ and $y$ - symmetryDefine symmetric relation R on set SProve that if $R$ is a symmetric, transitive relation on $A$ and the domain of $R$ is $A$, then $R$ is reflexive on $A$.If R is symmetric, must $S= (X,Y)in P(A)times P(A) $ be symmetric?Given set A, is the relation A x A always anti symmetric?Definition of symmetric relationName for relation property: If $xRy$ and $xRz$ and $x not =y$, then $yRz$.Can a relation be transitive when it is symmetric but not reflexive?













6












$begingroup$


Why is a symmetric relation defined by $forallxforally(xRy implies yRx)$ and not $forallxforally(xRy iff yRx)$?
(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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    "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
    yesterday






  • 2




    $begingroup$
    The definitions are equivalent.
    $endgroup$
    – PyRulez
    yesterday















6












$begingroup$


Why is a symmetric relation defined by $forallxforally(xRy implies yRx)$ and not $forallxforally(xRy iff yRx)$?
(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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    "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
    yesterday






  • 2




    $begingroup$
    The definitions are equivalent.
    $endgroup$
    – PyRulez
    yesterday













6












6








6


2



$begingroup$


Why is a symmetric relation defined by $forallxforally(xRy implies yRx)$ and not $forallxforally(xRy iff yRx)$?
(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.










share|cite|improve this question











$endgroup$




Why is a symmetric relation defined by $forallxforally(xRy implies yRx)$ and not $forallxforally(xRy iff yRx)$?
(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.







discrete-mathematics logic definition relations






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share|cite|improve this question













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share|cite|improve this question








edited yesterday









Asaf Karagila

307k33438770




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asked yesterday









Rodrigo SangoRodrigo Sango

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1326











  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    "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
    yesterday






  • 2




    $begingroup$
    The definitions are equivalent.
    $endgroup$
    – PyRulez
    yesterday
















  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    "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
    yesterday






  • 2




    $begingroup$
    The definitions are equivalent.
    $endgroup$
    – PyRulez
    yesterday















$begingroup$
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
yesterday




$begingroup$
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
yesterday




1




1




$begingroup$
"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
yesterday




$begingroup$
"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
yesterday




2




2




$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
yesterday




$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
yesterday










4 Answers
4






active

oldest

votes


















7












$begingroup$

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 $(x,y)in Rimplies (y,x)in R$ 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 $(1,2)in Rimplies (2,1)in R$ 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:



$(1,1)in Rimplies (1,1) in R$



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



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



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



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



With 2) we also have nine statements:



$(1,1)in Riff (1,1) in R$



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



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



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



...etc....



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



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






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Got it. Thank you, fleablood.
    $endgroup$
    – Rodrigo Sango
    yesterday






  • 1




    $begingroup$
    "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
    yesterday










  • $begingroup$
    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
    yesterday


















10












$begingroup$

If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$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 $Aimplies B$ than $Aiff B$. 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.






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.



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






    share|cite|improve this answer









    $endgroup$




















      1












      $begingroup$

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



      This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$



      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:$$aRbiff bRa$$



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



      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 that like that it is good custom to go for the one with less implications.






      share|cite|improve this answer









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        4 Answers
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        4 Answers
        4






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        active

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        7












        $begingroup$

        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 $(x,y)in Rimplies (y,x)in R$ 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 $(1,2)in Rimplies (2,1)in R$ 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:



        $(1,1)in Rimplies (1,1) in R$



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



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



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



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



        With 2) we also have nine statements:



        $(1,1)in Riff (1,1) in R$



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



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



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



        ...etc....



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



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






        share|cite|improve this answer









        $endgroup$












        • $begingroup$
          Got it. Thank you, fleablood.
          $endgroup$
          – Rodrigo Sango
          yesterday






        • 1




          $begingroup$
          "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
          yesterday










        • $begingroup$
          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
          yesterday















        7












        $begingroup$

        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 $(x,y)in Rimplies (y,x)in R$ 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 $(1,2)in Rimplies (2,1)in R$ 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:



        $(1,1)in Rimplies (1,1) in R$



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



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



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



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



        With 2) we also have nine statements:



        $(1,1)in Riff (1,1) in R$



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



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



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



        ...etc....



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



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






        share|cite|improve this answer









        $endgroup$












        • $begingroup$
          Got it. Thank you, fleablood.
          $endgroup$
          – Rodrigo Sango
          yesterday






        • 1




          $begingroup$
          "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
          yesterday










        • $begingroup$
          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
          yesterday













        7












        7








        7





        $begingroup$

        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 $(x,y)in Rimplies (y,x)in R$ 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 $(1,2)in Rimplies (2,1)in R$ 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:



        $(1,1)in Rimplies (1,1) in R$



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



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



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



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



        With 2) we also have nine statements:



        $(1,1)in Riff (1,1) in R$



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



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



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



        ...etc....



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



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






        share|cite|improve this answer









        $endgroup$



        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 $(x,y)in Rimplies (y,x)in R$ 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 $(1,2)in Rimplies (2,1)in R$ 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:



        $(1,1)in Rimplies (1,1) in R$



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



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



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



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



        With 2) we also have nine statements:



        $(1,1)in Riff (1,1) in R$



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



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



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



        ...etc....



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



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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        fleabloodfleablood

        73.1k22790




        73.1k22790











        • $begingroup$
          Got it. Thank you, fleablood.
          $endgroup$
          – Rodrigo Sango
          yesterday






        • 1




          $begingroup$
          "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
          yesterday










        • $begingroup$
          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
          yesterday
















        • $begingroup$
          Got it. Thank you, fleablood.
          $endgroup$
          – Rodrigo Sango
          yesterday






        • 1




          $begingroup$
          "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
          yesterday










        • $begingroup$
          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
          yesterday















        $begingroup$
        Got it. Thank you, fleablood.
        $endgroup$
        – Rodrigo Sango
        yesterday




        $begingroup$
        Got it. Thank you, fleablood.
        $endgroup$
        – Rodrigo Sango
        yesterday




        1




        1




        $begingroup$
        "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
        yesterday




        $begingroup$
        "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
        yesterday












        $begingroup$
        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
        yesterday




        $begingroup$
        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
        yesterday











        10












        $begingroup$

        If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$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 $Aimplies B$ than $Aiff B$. 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.






        share|cite|improve this answer









        $endgroup$

















          10












          $begingroup$

          If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$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 $Aimplies B$ than $Aiff B$. 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.






          share|cite|improve this answer









          $endgroup$















            10












            10








            10





            $begingroup$

            If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$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 $Aimplies B$ than $Aiff B$. 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.






            share|cite|improve this answer









            $endgroup$



            If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$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 $Aimplies B$ than $Aiff B$. 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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            José Carlos SantosJosé Carlos Santos

            170k23132238




            170k23132238





















                3












                $begingroup$

                If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.



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






                share|cite|improve this answer









                $endgroup$

















                  3












                  $begingroup$

                  If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.



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






                  share|cite|improve this answer









                  $endgroup$















                    3












                    3








                    3





                    $begingroup$

                    If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.



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






                    share|cite|improve this answer









                    $endgroup$



                    If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.



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







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    Viktor GlombikViktor Glombik

                    1,2372528




                    1,2372528





















                        1












                        $begingroup$

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



                        This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$



                        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:$$aRbiff bRa$$



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



                        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 that like that it is good custom to go for the one with less implications.






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

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



                          This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$



                          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:$$aRbiff bRa$$



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



                          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 that like that it is good custom to go for the one with less implications.






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

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



                            This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$



                            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:$$aRbiff bRa$$



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



                            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 that like that it is good custom to go for the one with less implications.






                            share|cite|improve this answer









                            $endgroup$



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



                            This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$



                            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:$$aRbiff bRa$$



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



                            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 that like that it is good custom to go for the one with less implications.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            drhabdrhab

                            103k545136




                            103k545136



























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