Originally Posted by

**dwsmith** Def:

R is irreflexive iff. $\displaystyle a\nsim a \ \forall a\in A$

Not irreflexive iff. $\displaystyle \exist a\in A \ni a\sim a$

R is asymmetric iff. if $\displaystyle a\sim b$, then $\displaystyle b\nsim a$

Not asymmetric iff. $\displaystyle \exist a\sim b$ and $\displaystyle b\sim a$

R is anti-symmetric iff. if $\displaystyle a\sim b \ \mbox{and} b\sim a$, then $\displaystyle a=b$.

Not anti-symmetric iff. $\displaystyle \exist a\sim b \ \mbox{and} \ b\sim a \ \mbox{and} \ a\neq b$

$\displaystyle \bigtriangleup ={(a,a)|a\in A}$

Proofs:

Let R be a nonempty relation on A. If R is symmetric and transitive, then R is not irreflexive.

If this one was true then there wouldn't be equivalence relations...(Itwasntme)

Tonio

Since R is symmetric, assume $\displaystyle a\sim b \ \mbox{and} \ b\sim a$. Then, since R is transitive, $\displaystyle a\sim a$. Therefore, R is not symmetric.

Q.E.D.

If $\displaystyle R\cap R^{-1}=\bigtriangleup$, then R is anti-symmetric.

(1)$\displaystyle R\cap R^{-1}\subseteq\bigtriangleup$

Let $\displaystyle (a,b)\in R\cap R^{-1}$. Therefore, $\displaystyle (a,b)\in R \ \mbox{and} \ (a,b)\in R^{-1}\leftrightarrow (b,a)\in R$. By the anti-symmetric hypothesis, if $\displaystyle a\sim b \ \mbox{and} \ b\sim a$, then $\displaystyle a=b$.

(2)$\displaystyle \bigtriangleup\subseteq R\cap R^{-1}$

$\displaystyle (a,a)\in\bigtriangleup\rightarrow a=a\rightarrow a=b\rightarrow a\sim b \ \mbox{and} \ b\sim a$$\displaystyle \rightarrow (a,b)\in R^{-1}$ Thus, $\displaystyle (a,b)\in R\cap R^{-1}$

Hence, R is anti-symmetric.

If R is transitive and irreflexive, then R is asymmetric.

By contradiction: If R is transitive and irreflexive, then R is not asymmetric.

Let $\displaystyle a\sim b \ \mbox{and} \ b\sim a$. By transitivity, $\displaystyle a\sim a$. However, R is irreflexive. Therefore, we have reached a contradiction and R is asymmetric.

Are these all logically correct?