(pq)$ which implies that $(tk)\in L$.
Hence a factor of $a$ is in $L:(pq)$.
It remains the case $(rs)<(pq)$. According to (b) we have
$(rs)\in J$. Since by assumption $J$ is linear, from the
arguments above it follows that $J$ is generated by all
$(uv):(ij)$ such that $(uv)\in I$ and $(uv),(ij)$ is not
a bad pair. Thus, there exists $(ml)\in I$
such that $(rs)\in (ml):(ij)$ and $(ml),(ij)$ is not a bad pair.
By Case 1 we may assume that $(ml)=(ir)$. Then if $k\neq j$,
we have $(rk)(ij)=(ir)(kj)\in I$, therefore $(rk)\in J$.
If $k=j$ then $(rq)(ij)=(ir)(qj)\in I$, so that $(rq)\in J\subset L$.
Since $(rj)(pq)=(pj)(rq)$, one obtains that $(rj)\in L:(pq)$.
This completes the proof of (1).
(2) We show first the following claim:
Let $(rs)\in L:(pq)$ and assume that $(rs)\notin L$. Then
we have the following:
(c) All the elements of the set $\{r,s,p,q\}$ are different.
(d) If $r~~(ps)\in J\subset L$,
$(ps)$ is not contained in any initial ideal in $L$.
Therefore there exists a quotient $L':(p'q')\subset L$ such that
$(ps)\in L':(p'q')$ and $(ps)\notin L'$ for some $(p'q')> (pq)$.
Since $(ps)\in J\subset L$ and $(ps)\notin L'$ one has $L'\neq L$,
so that by assumption $(p'p)\in L'$ or $(q'p)\in L'$.
First assume $(p'p)\in L'$. Then if $q\neq q'$, one has
$(pq)(p'q')=(p'p)(q'q)\in L'$, therefore we obtain the contradiction
$(pq)\in L':(p'q')\subset L$. Hence $q= q'$. But then $r\neq q'$,
so that $(pr)(p'q')=(p'p)(rq')\in L'$. Hence $(pr)\in L':(p'q')$
contradicting our assumption. Suppose now that $(q'p)\in L'$.
Then if $q\neq p'$, one has $(pq)(p'q')=(q'p)(qp')$, and if $q=p'$,
then $(pr)(p'q')=(q'p)(p'r)$.
In each case we get a contradiction which shows (d).
We will show that $L:(pq)$ satisfies the following condition $(*)$:
\begin{quote}
For every bad pair $(st)>(ml)$ in $L:(pq)$ at least one of the elements
$(sm),(sl),(tm),(tl)$ belongs to $L:(pq)$ and is bigger than $b$.
\end{quote}
\noindent
It is clear that this implies that $L:(pq)$ has linear quotients.
See \cite[Theorem 3.10]{AHH}.
If $L$ is an initial ideal then by \cite[Theorem 3.10]{AHH},
$L:(pq)$ satisfies the condition $(*)$, so we may assume that $J$
satisfies $(*)$.
Let $a=(st)>(ml)=b$ be a bad pair in $L:(pq)$.
Since $L:(pq)$ is linear, as we noted already, we have $a\in (uv):(pq)$
and $b\in (u'v'):(pq)$ with $(uv),(u'v')\in L$ such that $(uv),(pq)$
and $(u'v'),(pq)$ are not bad pairs. We may assume that $(uv)=(qs)$,
so that $t\neq p$. There are two cases:
$(u'v')=(pl)$ and $(u'v')=(ql)$. We will consider the first one,
the other one being treated similarly. So, assume $b\in (pl):(pq)$,
$(pl)\in L$. Since $(sl)(pq)=(pl)(qs)\in L$,
one obtains that $(sl)\in L:(pq)$, therefore if $(sl)>b$ we are done.
Hence we may assume $(sl)~~**b$,
one obtains $t b$.
If $t\neq q$, then $(tl)(pq)=(pl)(tq)\in L$, so that $(tl)$ satisfies
the desired condition. So, we may assume that $t=q$. Then $a\in L$.
If $m=p$ then $b\in L$ too. Since $(qp)> a>b$, by (b), $a,b$ is
a bad pair in $J$ and by assumption $J$ satisfies $(*)$.
Therefore $(ql)$ or $(ps)$ belongs to $J\subset L$.
Hence we may assume $m\neq p$. Then $(ms)(pq)=(qs)(mp)\in L$,
so if $(ms)>b$ we are done.
Thus it remains the case $a=(qs)\in L$, $q**