When are the braid relations in a quasitriangular Hopf algebra equivalent? Quasitriangular Hopf algebras have to satisfy, amongst other conditions, the following equations:
$$(\Delta \otimes \mathrm{id}) (R) = R_{13} R_{23}$$
$$(\mathrm{id} \otimes \Delta) (R) = R_{13} R_{12}$$
It appears that these two equations are not equivalent, although I don't know a single example for an $R$ that satisfies one of the equations and not the other.
Are there known additional conditions on the Hopf algebra that imply the equivalence of the two braid relations?
Or otherwise:
Is there a Hopf algebra $H$ and an element $R \in H \otimes H$ that satisfies e.g. the first braid relation all further axioms, but not the second braid relation?
Motivation: I think I have discovered a case where they are equivalent, which is the case of a quantum group with half-twist, as described in this article by Noah Snyder and Peter Tingley. It deals with Hopf algebras where $R = \left(t^{-1} \otimes t^{-1}\right) \Delta(t)$ for a suitable element $t$ in the Hopf algebra which is called the half-twist. I'm currently writing this particular instance up, but I'm very curious whether there are other conditions that make the braid relations equivalent, or counterexamples. It's nice if they are equivalent, since it saves you a lot of work when searching for new examples.
 A: The condition $R_{21}\,R=I$ (to be triangular) implies the equivalence of the two equations.
For the second question, let $A$ be a finite abelian group, and $H=k^A$ the Hopf algebra of function on $A$. Functions $f:A\times A\to \mathbb{C}^*$ (or equivalently maps from $A$ to $\operatorname{Maps}(A,\mathbb{C}^*$)) are in correspondence with invertible elements in $H^{\otimes 2}$. The first equation is equivalent to $f(a,b+c)=f(a,b)f(a,c)$ and similarly for the second one. Then a map $R: A\to \operatorname{Hom}(A,\mathbb C^*)$ that is not a group homomorphism and $R(0)=0$, works as an example.
A: Those conditions alone don't determine a quasitriangular structure.
An old paper of Radford's shows that the quasitriangular structure conditions for a finite dimensional Hopf algebra $H$ define Hopf algebra homomorphisms from $H^{*\text{cop}}\to H$.   This is obtained by considering the obvious linear injection $H\otimes H\to \operatorname{Hom}_{k}(H^*,H)$. 
Theorem 2.1 of this paper classifies all such Hopf morphisms in the case where $H=D(G)$ is the double of a finite group.  In Theorem 5.3 necessary and sufficient conditions are given for such a morphism to define a quasi-triangular structure.  Example 4.2 yields a class of Hopf morphisms that satisfy both of your conditions but do not yield quasitriangular structures.  It should be relatively easy to extract from Theorem 2.1 the conditions that control algebra versus coalgebra structures, and so help you construct examples which are algebra but not coalgebra morphisms or the other way around.  Possibly you may need a computer for this, but I can see no reason why such should not exist for some group $G$.

EDIT:
Actually, on further thought, we don't even need to go that far.  Consider $H=k^G$, the linear dual of the finite group $G$ with the usual Hopf algebra structure.  Then the coalgebra morphisms $f\colon H^{*\text{cop}}\to H$ are in bijective correspondence with set maps $f\colon G\to \widehat{G}$, where $\widehat{G}$ is the group of $k$-linear characters of $G$.  For such an $f$ setting
$$ R= \sum_{g\in G} e_g\otimes f(g),$$
then $f$ coalgebra is equivalent to the second braid relation $\operatorname{id}\otimes\Delta(R) = R_{13}R_{12},$ but $R$ satisfies the first braid relation if and only if $f$ is an (anti-)algebra map.  So you need only choose a set map $f\colon G\to\widehat{G}$ (sending the identity to the trivial character, even) which is not an algebra map.  As long as $\widehat{G}\neq 1$ this can always be done.
On the other hand (and dually, in this case), any algebra map $f\colon H^{*\text{cop}}\to H$ dualizes to a coalgebra map $kG\to k^{G}$, and we can again pick any set map $f\colon G\to \widehat{G}$ (preserving the identity) which is not an algebra map to find an element $R$ which satisfies the first braid relation but not the second.
