XSLT list item position issue - xslt

Question

pls check the below code. I am matching the list element and as per the below code the position() function returns correct number of the element with in ul elements i.e 1,2,3 where as if i do the test with position i get only ss,ss,ss . Can anyone let me know where am missing ?
XSL-FO
<xsl:template match="html:li" priority="2">
<fo:list-item>
...
<xsl:value-of select="position()"/> -- this returns 1,2,3
<xsl:if test="position() = 1">
<xsl:text>ss</xsl:text> -- only executes
</xsl:if>
<xsl:if test="position() = 2">
<xsl:text>ssddddd</xsl:text> -- does not only execute
</xsl:if>
<xsl:if test="position() = 3">
<xsl:text>sskkkkkkkkk</xsl:text> -- does not only execute
</xsl:if>
</fo:list-item>
</xsl:template
XSL
<xsl:template match="list">
<xsl:if test="list.item">
<xsl:variable name="styleAttr">
<xsl:text>margin-top: 1em;</xsl:text>
</xsl:variable>
<div>
<xsl:if test="string-length($styleAttr) > 0">
<xsl:attribute name="style">
<xsl:value-of select="$styleAttr"/>
</xsl:attribute>
</xsl:if>
<ul>
<xsl:apply-templates select="node()[not(self::list)]" />
</ul>
</div>
</xsl:if>
</xsl:template>
<xsl:template match="list.item" priority="1">
<li style="padding: 0;">
<div style="margin-bottom: 0.5em;">
<xsl:apply-templates />
<xsl:if test="following-sibling::node()[1][self::list]">
<xsl:apply-templates select="following-sibling::node()[1]" />
</xsl:if>
</div>
</li>
</xsl:template>