Generating implicit wrapper element

[Ferdinand Soethe <xml-dev@soethe.net>]

I have an XML-document with paragraphs and list items that have no
wrapper element around each list.

Something like this:

<par>my first para</par>
<par>second para</par>
<li>first list item of first list</li>
<li>second list item of first list</li>
<li>third list item of first list</li>
<par>third para</par>
<li>first list item of second list</li>
<par>fourth para</par>
<li>first list item of third list</li>
<li>second list item of third list</li>

In my transformation I would like to add these implicit wrapper
element around each of the list to get something like

<p>my first para</p>
<p>second para</p>
<ul>
<li>first list item of first list</li>
<li>second list item of first list</li>
<li>third list item of first list</li>
</ul>
<p>third para</p>
<ul>
<li>first list item of second list</li>
</ul>
<p>fourth para</p>
<ul>
<li>first list item of third list</li>
<li>second list item of third list</li>
</ul>

My attempt to solve this with

<xsl:template match="Aufzaehlungspunkt">
    
  <xsl:if test="not(preceding-sibling::li)">
    <ul>
  </xsl:if>
  
  <li><xsl:apply-templates/></li>

  <xsl:if test="not(following-sibling::li)">
    </ul>
  </xsl:if>

</xsl:template>

ran into two separate problems:

1. I'm not allowed to use <ul> without the matching </ul> (even though
   that is in a separate if-branch below.

2. The preceding-sibling and following-sibling-axes are only true when
   dealing with the very first and the very last <li> in the document.
   So rather than meaning 'previous element is not <li>' the meaning
   'there is no more previous <li>-element in this document'.

Any ideas how to solve these problems with XML?

Oh, btw. I saw the solution suggested some years ago

<officers>
<xsl:for-each select='/doc/person[@er="officer"]'>
<person><xsl:value-of select='.'/></person>
</xsl:for-each>
</officers>

<xsl:for-each select='/doc/person[not(@er="officer")]'>
<person><xsl:value-of select='.'/></person>
</xsl:for-each>

</perslist>
 </xsl:template>

 but this doesn't seem to be an option since I'm not using the
 procedural style of processing my document.

 Thanks for any input,

 --
 Ferdinand

--
Ferdinand Soethe