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April 21, 2005



Um...why is it that, if the relation is transitive and symmetric, it must be reflexive? I don't think that follows. Instead, siblinghood seems to be just the sort of relation that is symmetric, transitive, and not reflexive.


Ditto the last question. Im not my own brother, but brotherhood is symmetrical and transitive.


B and Rich,

Transitivity and symmetry jointly imply reflexivity. Let xRy be some dyadic relation over some set. Let a,b be members of that set. Suppose aRb. Then by symmetry, bRa. And note that by transitivity, if aRb and bRc, then aRc for arbitrary c in the set. Thus if aRb and bRa, then aRa. Generalising on a, relation R is reflexive .

Jeff Medina

I agree that it sounds odd to say someone is his or her own sibling, but it sounds markedly less odd to me when one substitutes the meaning of sibling you offered instead.
I.e., "has the same parents as" instead of "is the sibling of."


B and Rich,

Brotherhood is a transitive relation, but it is not symmetric. If I were a girl and had a male sibling, then I would have a brother. But surely this doesn't logically imply that I'm my male sibling's brother!


Jeff -
I was indicating a necessary condition for full siblinghood. The idea is that these considerations about how odd it is to say that one is his or her own sibling help us realise that there might be more to the necessary and sufficient conditions of standing in the relation than the one listed. On dictionary.com they say siblinghood is a relation stood in by two people when they share one or both parents. So one could say, for all x and y, x is a full sibling of y iff (i) x and y are distinct, and (ii) x and y have both parents in common. If the relation is cashed out in this way, it can't be transitive. But do you guys think this is the right way to cash out the relation?


Alex, D'oh... damn hangovers...but...

'is a full sibling of' means 'is a full brother or full sister of' right? The set of my siblings is just the union of the set of my brothers and the set of my sisters. So im a member of that set iff i am either my own brother or my own sister. Which is just as weird.

Im inclined to think that the relation is non transitive, but is 'pseudo-transitive' for want of a better word. That is, if Rab & Rbc, then Rac, unless a=c. Having tried out the straw poll test, everyone seems to think that 'x is a sibling of y' means something like 'x is a brother or sister of y, but is not y'.

Either that or Jeffs right and we just don't realise that 'is a sibling of' means 'has the same parents as'.

Jeff Medina


I had no intention of asserting that 'is a sibling of' means 'has the same parents as.' Shieva mentioned it as a necessary (but not necessarily sufficient) condition, so I worked from there and suggested a possible view of the relation that allowed for me to be my own sibling with a minimum of strangeness.

While I'm fine with positing such a thing philosophically, it seems clear to me that colloquial usage requires the intransitive qualification Shieva noted. xSIBy & ySIBz -> xSIBz iff x!=z.


I got set this question in a logic class last year. Every logician I've spoken to about it had the same answer - siblinghood's not really transitive. Though I have to say that it's still hard to see it.


Jeff, sorry, i think im just having one of those days. All i meant to suggest was that the 'has the same parents as' would be the other way to go if you didnt want to play the 'nearly but not really transitive' card. But it does seem right that colloquial usage does require us to play that card.


I guess I don't see the big problem with just denying that siblinghood is transitive. It seems that the following is a clear counterexample: My brother is my sibling. I am my brother's sibling. But I'm not my own sibling. So, siblinghood is not transitive.

We think it's transitive perhaps because of these issues about non-identity, but we're just wrong.


I agree that denying that siblinghood is transitive is the way to go. But the proof sketch for transitivity and symmetry implying reflexivity makes crucial use of the assumption that aRb, where xRy is some dyadic relation over a set that has a, b as members. The proof fails in models with empty domains or in ones in which no point "sees" any other. In such models, transitivity and symmetry are vacuously satisfied, but reflexivity is not. This is what we should expect, since schemas for transitivity and symmetry are conditionals, while reflexivity is not. So they are satisfied whenever nothing satisfies their antecedents.


Heh, came across this while looking for answers to my Logic and Reasoning course, on which I have an exam tomorrow.
I now have the answer! Thanks :)


Hello. I just bumped into this page. Interesting thoughts. BUT: reflexivity is not implied by symmetry and transitivity. The empty relation on a non-empty set is not reflexive, but is symmetric and transitive.


You are correct. transitivity and symmetry do not imply reflexivity.

If you have an x such that xRy is false for all y, then xRz does not follow for that value of x and thus reflexivity is not guaranteed as a relation is only reflexive if xRx for all x in Omega.

Consider Omega = naturals and xRy iff x=y=0
This relation is symmetric and transitive, however if you take x=1, xRy is never true, thus reflexivity is not guaranteed, and indeed this relation is not reflexive.

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