{\bf An $\infty$-dimensional DD Theorem, take 1.}

Suppose $\bbA=\calU(\fraka)$ and $\bbB=\calU(\frakb)$ are Hopf algebras with their native products and with
$aS$, $a\Delta$, $bS$, $b\Delta$, etc. Suppose $\langle\cdot,\cdot\rangle\colon\frakb\otimes\fraka\to\bbQ$ is a
pairing such that:
\begin{itemize}
\item Compatibility of $[]_b$ and $a\Delta$ etc.
\item Non degeneracy.
\end{itemize}

Then
\begin{enumerate}
\item $\langle\cdot,\cdot\rangle$ extends uniquely to a non-degenerate pairing $\bbB\otimes\bbA\to\bbQ$ such that
$m$ and $\Delta$ are compatible.
\item $\bbD=\bbB\otimes\bbA$ is a Hopf algebra with the DD formulas and $\bbA\to\bbD$ and $\bbB\to\bbD$ are
Hopf morphisms.
\item If $b_i$ and $a_i$ are dual bases of $\bbB$ and $\bbA$ relative to our pairing, then $R=\sum b_\otimes
a_i$ satisfies the quasi-triangularity axioms.
\end{enumerate}

{\bf An $\infty$-dimensional DD Theorem, take 2.}

{\bf Step 1.} Everybody knows that if $H$ is a finite dimensional Hopf algebra then $D=H^\ast\otimes$ is a
quasi-triangular Hopf algebra, with $R$, $m$, $\Delta$, $S$ given by the following formulas\ldots

{\bf Step 2.} If $A$ and $B$ are Hopf algebras over a ring $\Omega$
with a Hopf pairing $P\colon A\otimes B\to\Omega$ and $R\in B\otimes A$ contracts $P$ to the identity, the same
conclusion holds for $D=B\otimes A$.

{\bf Step 3.} Over $\Omega=\bbQ\llbracket\hbar\rrbracket$
let $\bbA=\calU(\fraka)\llbracket\hbar\rrbracket$,
$\bbA'=\langle\hbar\fraka\rangle\subset\bbA$, and
$\bbB=\calU(\frakb)$, with $P\colon\bbA'\otimes\bbB\to\Omega$ be
given by $\langle \hbar a,b\rangle=\langle\hbar x,y\rangle=1$, and let
$R=\sum_{m,n}y^nb^m\otimes(\hbar a)^m(\hbar x)^n/m![n]_q!$. Then we're
in the situation of Step 2, with $A=\bbA'$ and $B=\bbB$, and hence
$\bbD'=\bbB\otimes\bbA'$ is a quasi-triangular Hopf algebra.

{\bf Step 4.} All the formulas extend $\Omega$-linearily to $\bbD=\bbB\otimes\bbA$ and hence all identities
hold there too.