The Wikipedia definition of an Antisymmetric relation

Question

The Wikipedia definition of an Antisymmetric relation says :

R is antisymmetric precisely if for all $a$ and $b$ in $X$

\begin{align} \text{if } R(a,b) \text{ and } R(b,a) \text{ then } a=b. \end{align}

or, equivalently

\begin{align} \text{if } R(a,b) \text{ with } a\ne b \text{ then } R(b,a) \text{ must not hold. } \end{align}

My Question : Shouldn't the contrapositive of the first statement say

\begin{align} \text{if } a\ne b \text{ then } R(a,b) \text{ or } R(b,a). \end{align}

and in considering the mathematical sense of "or", it might as well happen that even if $R(a,b)$ holds ; nothing stops from $R(b,a)$ NOT being true.

Please help me resolve this.

Answer 1

The contrapositive should be $$\text{if } a\ne b \text{ then it not true that } R(a,b) \text{ and } R(b,a)$$ and that is equivalent to $$\text{if } a\ne b \text{ then } R(a,b) \text{ does not hold or } R(b,a) \text{ does not hold }$$ which implies $$\text{if } R(a,b) \text{ with } a\ne b \text{ then } R(b,a) \text{ must not hold. }$$

Answer 2

No.

Let $R=\varnothing\subseteq X\times X$.

Then $R$ is antisymmetric but from $a\neq b$ it cannot be concluded that $\langle a,b\rangle\in R$ or $\langle b,a\rangle\in R$.

Answer 3

Precisely because "$p$ or $q$" involves the possibility that both $p$ and $q$ be true, your proposal cannot work, since if both $aRb$ and $bRa$, then you would have $a=b$, absurd.

Besides that, remember that not every pair $(a,b)$ must be such that at least one of $aRb$ or $bRa$ holds: that might help you understand the quoted definition.

Answer 4

The notion of contrapositive can be generalized in the following way:

Theorem: The following are logically equivalent:

• If $A$ and $B$ then $C$
• If $A$ and not $C$ then not $B$

This can be proven by using the following chain of equivalences

• $(A \wedge B) \implies C$
• $A \implies (B \implies C)$
• $A \implies (\neg C \implies \neg B)$
• $(A \wedge \neg C) \implies \neg B$

Answer 5

If your proposed condition was equivalent to antisymmetry then all partial orders would be total orders. Thus any non-total partial order provides a counterexample. For example, on the set of positive integers let $R(a,b)$ denote the relation $a$ divides $b$. It is easy to see that $R$ is antisymmetric and indeed a partial order. Note that $2 \neq 3$, but neither $R(2,3)$ nor $R(3,2)$ hold.