Definitions of enriched monoidal category This question is about two definitions of enriched monoidal categories I have:
Let $\mathcal{V}$ be a symmetric monoidal closed category.
The first definition: a $\mathcal{V}$-enriched category $\mathcal{C}$ is a pseudomonoid object in the Day-convolution monoidal category $(\mathcal{V}\text{-}\mathbf{Cat}, \otimes_{\mathrm{Day}}, y(1))$. This is a straightforward generalization of the definition of monoidal categories but everything has to be worked externally and I don't really feel like to work in this definition (for example, I am pretty sure that a kind of coherence theorem holds but I don't know if I will be able to write down the proof).
The second definition can be found here (Bar constructions for topological operads and the Goodwillie derivatives of the identity, Definition 1.10): https://projecteuclid.org/download/pdf_1/euclid.gt/1513799607
In this paper, an enriched monoidal category $(\mathcal{C}, \overline{\wedge}, S)$ over $(\mathcal{V}, \wedge, I)$ is a $\mathcal{V}$-enriched category $\mathcal{C}$ which is tensored (denoted by $\otimes$) and cotensored, together with a monoidal structure on the underlying category $\mathcal{C}_0$ and the distribution natural transformation $d: (X\wedge Y)\otimes (C \overline{\wedge} D)\to (X\otimes C)\overline{\wedge}(Y\otimes D)$ satisfying certain associativity and unitality condition, which is a natural requirement for bar constructions.
So my question is: What is the relation between these two? It will be pleasing to be able to replace the first definition by the second one in other situations but I could not see any immediate implication from one to another or confluence under some conditions. 
 A: The two definitions are equivalent, for monoidal structures on $\mathcal{V}$-categories that are tensored over $\mathcal{V}$. I'll describe how the tensor product corresponds to the distributivity map: 
The tensor product for a monoidal $\mathcal{V}$-category $\mathcal{C}$ in the first sense is given by specifying a tensor product on objects, and compatible maps $\mathcal{C}(C,D) \wedge \mathcal{C}(C',D') \to \mathcal{C}(C\overline{\wedge}C', D\overline{\wedge} D')$ in $\mathcal{V}$. So to show the second definition implies the first, we need to construct these maps. If $\mathcal{C}$ is tensored over $\mathcal{V}$, we can use the tensor product in $\mathcal{C}_0$ to get a natural map of sets
$$ \mathcal{V}(X, \mathcal{C}(C,D)) \times \mathcal{V}(Y, \mathcal{C}(C',D')) \cong \mathcal{C}_0(X \otimes C, D) \times \mathcal{C}_0(Y \otimes C', D') \to \mathcal{C}_0((X \otimes C) \overline{\wedge} (Y \otimes C'), D \overline{\wedge} D').$$
Using the distributivity transformation we get a map
$$ \mathcal{C}_0((X \otimes C) \overline{\wedge} (Y \otimes C'), D \overline{\wedge} D') \to \mathcal{C}_0((X \wedge Y) \otimes (C \overline{\wedge} C'), D \overline{\wedge} D') \cong \mathcal{V}(X \wedge Y, \mathcal{C}(C \overline{\wedge} C', D \overline{\wedge} D').$$ Now set $X = \mathcal{C}(C,D)$ and $Y = \mathcal{C}(C',D')$, then the pair of identity maps gets sent to the required morphism $\mathcal{C}(C,D) \wedge \mathcal{C}(C',D') \to \mathcal{C}(C\overline{\wedge}C', D\overline{\wedge} D')$.
On the other hand, if we have these maps then for $X,Y \in \mathcal{V}$, $C,D \in \mathcal{C}$, we can form the composite
$$ X \wedge Y \to \mathcal{C}(C, X \otimes C) \wedge \mathcal{C}(D, Y \otimes D) \to \mathcal{C}(C \overline{\wedge} D, (X \otimes C) \overline{\wedge} (Y \otimes D)),$$
where the first map is the tensor product of unit maps for the cotensoring adjunction. This is then adjoint to the distributivity morphism $(X \wedge Y) \otimes (C \overline{\wedge} D) \to (X \otimes C) \overline{\wedge} (Y \otimes D))$.
