in SAT resolution proofs, are all DAGs possible?
Let's assume that you do not allow Tautologies in the leafs, and consider a structure where exactly 3 leafs have edges to a node $A$. Denote the leaf clauses by $(a\vee L_1),(\neg a\vee b\vee L_2),(\neg b\vee L_3)$, where $L_1,L_2,L_3$ are discunctions of literals.
Since they all have edges to $A$, we have the following:
Leaves 1,2 resolve to $(b\vee L_1\vee L_2)$.
Leaves 2,3 resolve to $(\neg a\vee L_2\vee L_3)$

Since $A$ is a single node, then these expressions are equal. 
Since there is no initial tautology, then $b\notin L_3$, but this implies that $b\in L_2$, which isn't the case (assume we have no duplicates).

So the only possibility is $b=\neg a$, but this means that $L_1=L_2\vee L_3$ and $L_3=L_2\vee L_3$, so the leaves aren't distinct.

If I am not mistaken, then this means that this structure is not possible, so not every DAG is possible (under the assumptions).