Verify that $R$ is the cover relation of a partially ordered set. I am recently learning about Partial Ordering and for some reason, I am finding it difficult to grasp all the ideas involved in it. 
Problem:  Let $X=\{a,b,c,d,e,f\}$ and let the relation $R$ on $X$ be defined by $aRb$, $bRc$, $cRd$, $aRe$, $eRf$, $fRd$. Verify that $R$ is the cover relation of a partially ordered set, and determine all linear extensions of this partial order. 
My Attempt: I picked up how to draw a Hasse diagram and so this is the picture I drew: 

I am guessing that the diagram is somehow involved in the proof, but I am not sure. Any hints on how to proceed after this step will be much appreciated. 
 A: I think taking the reflexive and transitive closure of $R$ should do it, but it's a tedious check. So define $S$ on the same set by $aSa, bSb, cSc, dSd ,eSe, fSf, aSb, bSc, aSc, cSd, aSd, aSe, eSf,aSf, fSd, eSd, bSd$ (I think these are all) and check that $S$ is a partial order and then check that $xSy$ and no $z$ with $xSz$ and $zSy$ implies $xRy$ and vice versa. It's almost trivial from the picture and construction, really. There is probably a criterion for $R$ to be the covering relation of a poset. 
The linear extensions are distinguished by how we order $b,c,e,f$ in a linear way, respecting $b < c$ and $e< f$. Options are $ b < c < e < f$, $b < e < c < f$, $b < e < f < c$ and $e < f < b < c$ and $e < b < f < c$, $e < b < c < f$ as far as I can see, so 6 options.
A: In this context it is handsome to make use of the concept strict partial order, which is by definition a relation that is irreflexive and transitive. If you define $\overline R$ as the transitive closure of $R$ then  it is not difficult to verify that in your case $\overline R$ is irreflexive. Also $\overline R$ is transitive so it is a strict partial order. Just like a partial order it has a covering relation $C$ and evidently we have $C\subseteq R$. It remains to be shown that also $R\subseteq C$, and this can be done by brute force. The covering relation of strict partial order $\overline R$ is evidently the same as the covering relation of the corresponding partial order $\overline R\cup\triangle$, where $\triangle:=\{\langle x,x\rangle\mid x\in X\}$.

In general:


*

*If $P$ is a strict partial order then $P\cup\triangle$ is a partial order.

*If $P$ is a partial order then $P\setminus\triangle$ is a strict partial order.

