janitorial: simplify test expression without semantics change

The "sComposerFilter != m_sRowSetFilter" could not influence the result.
Proof.
 The right hand-side of the || is evaluated only when m_sRowSetFilter.isEmpty()
 So the only case where the removed test could have an influence is when
 m_sRowSetFilter.isEmpty().
 However, the right hand-side of the && is evaluated only when !sComposerFilter.isEmpty().
 We have m_sRowSetFilter.isEmpty() and !sComposerFilter.isEmpty().
 In particular, (sComposerFilter != m_sRowSetFilter) is true.
Qed.

Change-Id: I1484d78cf2d7a5e8ca44f382eb7c440c84d8c10e

dbaccess/source/core/api/OptimisticSet.cxx

@@ -157,7 +157,7 @@ void OptimisticSet::makeNewStatement( )
     fillJoinedColumns_throw(m_aSqlIterator.getJoinConditions());
 
     const ::rtl::OUString sComposerFilter = m_xComposer->getFilter();
-    if ( !m_sRowSetFilter.isEmpty() || (!sComposerFilter.isEmpty() && sComposerFilter != m_sRowSetFilter) )
+    if ( !m_sRowSetFilter.isEmpty() || !sComposerFilter.isEmpty() )
     {
         FilterCreator aFilterCreator;
         if ( !sComposerFilter.isEmpty() && sComposerFilter != m_sRowSetFilter )