Non-degeneracy of comultiplication (multiplier Hopf algebras) Consider the following fragment from the paper "Multiplier Hopf-algebras" by Van Daele.

Can someone explain how the coassociativity in definition 2.2 (ii) and the requirement $(\Delta \otimes \iota)\Delta = (\iota \otimes \Delta)\Delta$ are equivalent? I understand how we can extend these morphisms to the multiplier algebra.
I can see that if $(\Delta\otimes \iota)\Delta = (\iota \otimes \Delta)\Delta$, then
\begin{align}&(a \otimes 1 \otimes 1)(\Delta \otimes \iota)(\Delta(b) (1 \otimes c)) \\
&= (a \otimes 1 \otimes 1)(\Delta \otimes \iota)(\Delta(b))(1 \otimes 1 \otimes c) \\
&= (a \otimes 1 \otimes 1)(\iota\otimes \Delta)(\Delta(b)) (1\otimes 1 \otimes c)\\
&= (\iota \otimes \Delta)((a \otimes 1)\Delta(b))(1 \otimes 1 \otimes c)\end{align}
But what about the converse?
 A: As in your other question let's unpack this, and be careful.  Set
$$ \alpha = \Delta\otimes\iota, \quad \beta = \iota\otimes\Delta, $$
which are homomorphisms $A\otimes A\rightarrow M(A\otimes A\otimes A)$.
These are nondegenerate and so have extensions $\overline\alpha, \overline\beta$ to $M(A\otimes A)$.  For example, we have by definition that
$$ \overline\alpha(x)\alpha(u)(a\otimes b\otimes c) = \alpha(xu)(a\otimes b\otimes c)
\qquad (x\in M(A\otimes A), u\in A\otimes A, a,b,c\in A). $$
Similarly "on the left".
Condition (ii) says that
$$ (a\otimes 1\otimes 1)\alpha(\Delta(b)(1\otimes c))
= \beta((a\otimes 1)\Delta(b))(1\otimes 1\otimes c). $$
Multiply this on the right by $\Delta(d)(e\otimes f)\otimes gh = \alpha(d\otimes g)(e\otimes f\otimes h)$ to get
$$ (a\otimes 1\otimes 1)\alpha(\Delta(b)(1\otimes c))\alpha(d\otimes g)(e\otimes f\otimes h)
= \beta((a\otimes 1)\Delta(b))(1\otimes 1\otimes c)\alpha(d\otimes g)(e\otimes f\otimes h), $$
which gives
$$ (a\otimes 1\otimes 1)\alpha(\Delta(b)(d\otimes cg))(e\otimes f\otimes h)
= \beta((a\otimes 1)\Delta(b))\alpha(d\otimes cg)(e\otimes f\otimes h). $$
This in turn shows
$$ (a\otimes 1\otimes 1)\overline\alpha(\Delta(b))\alpha(d\otimes cg))(e\otimes f\otimes h)
= \beta((a\otimes 1)\Delta(b))\alpha(d\otimes cg)(e\otimes f\otimes h). $$
Thus $(a\otimes 1\otimes 1)\overline\alpha(\Delta(b)) = \beta((a\otimes 1)\Delta(b))$.
Now multiply on the left by $ij\otimes (k\otimes l)\Delta(m) = (i\otimes k\otimes l)\beta(j\otimes m)$ to get
\begin{align*} (ija\otimes (k\otimes l)\Delta(m))\overline\alpha(\Delta(b)) &= (i\otimes k\otimes l)\beta(j\otimes m)\beta((a\otimes 1)\Delta(b)) \\
&= (i\otimes k\otimes l)\beta((ja\otimes m)\Delta(b)) \\
&= (i\otimes k\otimes l)\beta(ja\otimes m) \overline\beta(\Delta(b)) \\
&= (ija\otimes (k\otimes l)\Delta(m)) \overline\beta(\Delta(b)).
\end{align*}
Thus $\overline\alpha \circ \Delta = \overline\beta \circ \Delta$, which is what we wanted to prove.
There is a slightly quicker way to write this out, because we could just use that $\overline\alpha(x)\alpha(u) = \alpha(xu)$, without multiplying by $a\otimes b\otimes c$.

My understanding is that van Daele was originally motivated by the $C^*$-algebraic approach to Quantum Groups.  In this setting, you always have approximate identities, and this can somewhat simplify these sorts of arguments.  I believe that for some multiplier Hopf algebras (maybe "regular" ones?) you have "local units" which can be used similarly: of course, you need to prove that these exist.  In some later papers by van Daele and coauthors (perhaps some on the arXiv only) there are various discussions about how to efficiently perform calculations with multiplier Hopf algebras (the idea of "covering elements").  It might be interesting to study these.
