Studying relation properties. My definition of a transitive relation is as follows:

A relation is transitive if and only if $\forall a,b,c \in A [aRb \land bRc \implies aRc]$

My question is: if $aRb \land bRc$ never occurs in the first place, is the relation considered transitive?

I ask this because when I read $\forall a,b,c \in A [aRb \land bRc \implies aRc]$ I read it as follows: "when $a$ relates with $b$ and $b$ relates with $c$, $a$ relates to $c$". But since this "when" never happens, there is no condition to evaluate to decide if it is transitive or not.

The same issue occurs with antisymmetry, where $\forall a,b \in A[aRb \land bRa \implies a = b]$ - what if $aRb \land bRa$ never occurs in the first place?


I just remembered that $F_0\implies whatever$ is always $V_0$.... I guess this answers my question... kinda.

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    $\begingroup$ Short answer: Yes. $\endgroup$ – ՃՃՃ Dec 6 '12 at 1:32
  • $\begingroup$ You may want to look at Vacous truth. $\endgroup$ – ՃՃՃ Dec 6 '12 at 1:34

The answer is yes in both cases; these are examples of statements that are vacuously true, to use the usual expression. To see why we would want to use this interpretation, ask yourself what it would take in order to show that $R$ is not transitive: you’d need to find elements $a,b,c\in A$ such that $a~R~b$, $b~R~c$, and yet $a~\not R~c$. But if there are no elements $a,b,c\in A$ such that $a~R~b$ and $b~R~c$, you can’t possibly find such a counterexample. In short, there is no counterexample to transitivity of $R$, so by default $R$ must be transitive.

Another way to put it: transitivity says that when certain conditions are met, a certain thing has to happen. It says nothing about cases in which those conditions are never met: it requires nothing in those cases, so it cannot possibly fail.

  • $\begingroup$ If they ask me to verify if a relation is transitive or not and this scenario occurs, how should I word it out? Should I say "it is vacuously true"? $\endgroup$ – Zol Tun Kul Dec 6 '12 at 1:41
  • $\begingroup$ @Omega I would say it is. Or I would simply cite it as 'vacuous truth'. $\endgroup$ – 000 Dec 6 '12 at 1:48
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    $\begingroup$ @Omega: That works. ‘Is this relation transitive?’ ‘Yes, vacuously so.’ ‘Is it true that this relation is transitive?’ ‘Yes, it’s vacuously true’. $\endgroup$ – Brian M. Scott Dec 6 '12 at 2:28

Yes...it is...You can consider both properties to hold, unless there exists a counterexample:

Transitivity fails IF AND ONLY IF: $\quad \exists a, b, c \in A[a R b \land bRc \land \lnot (a R c)]$

Likewise for antisymmetry:

Antisymmetry fails IF AND ONLY IF: $\quad \exists a,b \in A[aRb \land bRa \land a \neq b]$


You have an if-then statement, $P\Rightarrow Q$. If $P$ is false, then the if-then statement is true. So, if you have a relation where $aRb\wedge bRc$ never occurs, it would be false to state $aRb\wedge bRc$.


On the set $\{-1,1\}$ with $R$ defined $xRy$ if $y=-x$, we see that $R$ is transitive by vacuity.

  • $\begingroup$ @downvoter I made an error in my original calculations. I've corrected this. $\endgroup$ – 000 Dec 6 '12 at 11:34

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