Justify a relation

let $\mathrm A = \Bbb Z \text{ and } R = \{(a,b) \in \mathrm A\times \mathrm A | a \lt b \}$ investigate whether the relationship is symmetric or antisymmetric.

So (...)

• Symetric [FALSE]: $(a,b) \in R \implies a \lt b \implies (b,a) \not\in R \text { as if } a\lt b \implies b \not\lt a$
• Antisymetric [TRUE]: $(a,b) \in R \land (b,a)\not\in R \implies a = b$.

Since if $P \land Q \implies S$, if $P$ or $Q$ is false, then no matter the logic value in $Q$, because $F \implies F \equiv \mathrm T$ and $F \implies V \equiv \mathrm T$, but I don't know how to justify this antisymetry in this relation.

There is a way to write this as a math notation?

Thanks!

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We want to show: $\forall x,y \in A\,(\,(x,y)\in R \land (y,x)\in R \implies x=y)$.

This is how I will do it:

Suppose $x$ and $y$ are arbitrary elements of $A$ and suppose $(x,y)\in R$ and $(y,x)\in R$. Then $x<y$ and $y<x$, but this is impossible. So, our statement is vacuously true.

Another way would be to prove the contrapositive: $\forall x,y \in A\,(\, x\ne y\implies (x,y)\not\in R \lor (y,x)\not\in R)$ which isn't difficult to prove.

Or you may directly write it out like this,

\begin{align} &\forall x,y \in A\,(\,(x,y)\in R \land (y,x)\in R \implies x=y)\\ &\equiv \forall x,y \in A\,(\,x<y \land y<x \implies x=y)\\ &\equiv \forall x,y \in A\,(\text{ False} \implies x=y)\\ &\equiv \forall x,y \in A\,(\text{ True })\\ \end{align} which is equivalent to $\text{True}$.

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What you have could stand to be polished up a bit, but it’s basically fine (though you’ve done more than necessary for symmetry). Here’s how I might write it:

• Symmetry: $R$ is not symmetric, because for instance $\langle 1,2\rangle\in R$, but $\langle 2,1\rangle\notin R$: $1<2$, but $2\not<1$.

• Antisymmetry: $R$ is antisymmetric: the antecedent of the implication $$\Big(\langle a,b\rangle\in R\land\langle b,a\rangle\in R\Big)\to a=b$$ is always false, so the implication is vacuously true.

Under symmetry you actually proved that $R$ is asymmetric: if $\langle a,b\rangle\in R$, then $\langle b,a\rangle\notin R$. You could write this up a little better as follows:

• Asymmetry: Suppose that $\langle a,b\rangle\in R$. Then $a<b$, so $b\not< a$, and therefore $\langle b,a\rangle\notin R$.

Note that the argument that you and I both gave for antisymmetry really shows that an asymmetric relation is always antisymmetric.

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