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There are two questions

S is the set of all people and R is the relation on set S such that $(a, b) \in R$, where $a$ and $b$ are people if $a$ is taller than $b$

According to the solution, this relation is anti-symmetric, which makes sense at first look. Clearly if $a$ is taller than $b$, then $b$ cannot be taller than $a$. Moving on to the second question;

S is the set of all people and R is the relation on set S such that $(a, b) \in R$, where $a$ and $b$ are people if $a$ is not taller than $b$

Now according to the solution, this is not anti-symmetric, since while it is true that if $a$ is not taller than $b$, then $b$ is not taller than $a$, unless $a$ and $b$ are of the same height, but that does not be $a = b$ since they are two distinct people, hence the equality does not hold.

The equality part is confusing me. If we look at the first question, and the definition of anti-symmetry

the relation R is antisymmetric if $aRb$ and $bRa$ implies $a = b$.

and clearly, $a$ cannot be taller than $a$, so why is the first question anti-symmetric while the second one is declared anti-symmetric based on the same equality technicality?

For more clarification, let $aRb$ in the first question. Then $b \not R a$ unless $a = b$ according to the very definition of anti-symmetry. But it is not true that $a R a$, so how can we declare it to be anti-symmetric?

Asaf Karagila
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Aditya
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    The first relation never fulfills the antecedent, so the implication is always true. – Damian Oct 21 '22 at 05:42
  • An equivalent definition of antisymmetry is: if $a\ne b$, then we can’t have both $aRb$ and $bRa$. Perhaps this is easier to consider for you? – Milten Oct 21 '22 at 06:02
  • @Damian ok that makes sense, thanks for clarifying – Aditya Oct 21 '22 at 07:04

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Logically, the comments by Damian and Milten are good explanations. $a\Rightarrow b$ is true if $a$ is always false, and conversely, $\neg b \Rightarrow \neg a$ is also true if $a$ is always false. Grasping this intuitively does require a little bit of brain gymnastics at first though.

It might be a little easier to get your head around it if we phrase things less mathematically precise and more in line with how we usually think about relations: an operator with something on either side of it. Instead of

the relation $R$ is antisymmetric if $aRb$ and $bRa$ implies $a=b$.

let's say

the relation is antisymmetric if you can never swap the two sides unless they're identical

or even more obviously

the relation is antisymmetric if you can never swap the two sides, possibly unless they're identical

Antisymmetry poses no requirement on $aRa$ -- it might be true, it might be false. If it is true, then yes, you can of course swap the two sides (because it doesn't really change anything).

But that's the only case where you would ever be allowed to do it. If the two sides aren't the same, no swapping.

ij7
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