What should I call a poset with the property that each element has AT MOST ONE 
 predecessor? 


(I'm actually interested in the special case in which there are no infinite descending chains.)


CLARIFICATION: I mean each element has a unique immediate predecessor.  

As you go higher up there can be branching, but never collapsing. That is, I don't want to allow $a < z$ and $b < z$ with $a$ and $b$ incomparable; but I am happy with 
$a< y$ and $a< z$ with $y$ and $z$ incomparable.


FURTHER: 


1. The immediate predecessor should be the largest predecessor. If I write 
$i-1$ for the unique immediate predecessor of $i$, then $i-1 \leq x \leq i$
forces either $x = i-1$ or $x = i$.
2. I should have said -- and now do say -- that each element has at most one immediate predecessor.



Terminology for posets.
A partial order with no infinite descending chains is said to be *well-founded*. Every well-founded partial order admits an ordinal ranking function, an assignment of nodes in the order to ordinals, such that higher nodes get larger ordinals. One can therefore speak of the ordinal rank of a node or the height of the whole order. It is common to prove things about such orders using transfinite induction, either on the ordinal ranks or on the partial order directly.


Your other property, about every node having a unique predecessor (do you mean immediate predecessor?), is a discreteness property, which by iterating is clearly incompatible with well-foundedness. 


But perhaps you mean that the partial order is well-founded, and every node has at most one immediate predecessor? In this case, the height of the order will be $\omega$, since there will be no node of limit rank. In this case, your partial order is: a well-founded partial order of height at most $\omega$.


But it still depends on what you mean by predecessor, since one could have a node $b$ with an immediate predecessor $a$, and also with an increasing sequence below $b$ to the side of $a$. Would this count for you? If so, then infinite ranks would still be possible.


(The usual meaning of predecessor in a partial order is simply that $a$ is a predecessor of $b$ if and only if $a\lt b$. With this meaning, the only partial order in which every element has a unique predecessor is the empty order, since if $z$ is a node in the order, it has a unique predecessor $y$, which also must have a predecessor $x$, which would also be a predecessor of $z$, a contradiction.)


