Please help solve following:
Let $P$ be the set of all people who have ever lived and define a relation $R$ on $P$ as follows: $\forall\; r,s \in P,\; r\,R\,s \iff r$ is an ancestor of $s$ or $r=s$.
Is $R$ a partial order relation? Prove or give a counter- example.
I think it is partial order based on following calculation, not sure if its correct though:
Reflexive: $(r,s)$ belong to $P, r\,R\,s$ because $r=r$ and $s=s$
Antisymetric: Then either $r=s$ or $r < s$; or $s=r$ 0r $s < r$; Thus $r\leq s$ or $s\leq r$. Hence $r=s$ is antisymetric
Transitive: Suppose $r$, $s$, $e$ are ordered such that $r\,R\,s$ and $s\,R\,e$; Then either $r=s$ or $r < s$ and either $s=e$ or $s < e$.
It follows that:
- $(r < s, s < e)$ by transitivity, $r < e$ and $r\,R\,e$;
- $(r < s, s=e)$ by substitution, $r < r$ and $r\,R\,e$;
- $(r=s, s < e)$ by substitution, $r < e$ and $r\,R\,e$;
- $(r=s, s=e)$ by definition of $R$, $r=s$, $s=e$, then $r=e$ and $r\,R\,e$;
In each case $r\,R\,e$, therefore $R$ is transitive. $R$ is partial order.
Reflexivity: $\mathfrak{R}$ is reflexive if $\forall r\in P,\,r\,\mathfrak{R}\,r$ which isn't what you did. In fact, if you take $r\in\mathfrak{R}$, $r=r$ so $r\,\mathfrak{R}\,r $ according to the definition.
Antisymetry: Yes that's right, even if in fact seems to me that you're using antisymetry of numbers. I think it's better to say "$r$ is ancestor of $s$ and $s$ ancestor or $r$ so they are the same person $r=s$ because if not, $r$ ancestor of $s$ means he's older than $s$ and $s$ ancestor of $r$ contradiction". So $r=s$.
Transitivity: Same remark for transivity. Don't forget elements of $P$ are humans not numbers.