RE: not(not() and not())

Gustaf,

DeMorgan's Law states:

  A & B => ¬(¬A | ¬B)

and

  A | B => ¬(¬A & ¬B)

(Where ¬ is the boolean 'not' operator).

That is, if you take two expressions A and B (such as
parent::tr/following-sibling::tr and
ancestor::tbody/following-sibling::tbody from your example),
then 

parent::tr/following-sibling::tr and
ancestor::tbody/following-sibling::tbody

is the same as

not(not(parent::tr/following-sibling::tr) or
not(ancestor::tbody/following-sibling::tbody))

NOTE: the 'and' has changed to an 'or'.

Regards,

Dan.

-- 
Danny Yates
Technical Architect
Abbey National Treasury Services
E-mail: Danny.Yates@xxxxxxxxxx
Phone: +44 20 7756 5012
Fax: +44 20 7612 4342


-----Original Message-----
From: Gustaf Liljegren [mailto:gustafl@xxxxxxxxxx]
Sent: 17 November 2002 04:15
To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx
Subject:  not(not() and not())


I seem to have a problem with logic. Can someone explain why this XPath
expression:

<!-- Paint border unless it's the last row in the last rowgroup. -->
<xsl:if test="not(not(parent::tr/following-sibling::tr) and
              not(ancestor::tbody/following-sibling::tbody))">
  <xsl:call-template name="draw-border-bottom"/>
</xsl:if>

is not exactly the same as this:

<!-- Paint border unless it's the last row in the last rowgroup. -->
<xsl:if test="parent::tr/following-sibling::tr and
              ancestor::tbody/following-sibling::tbody">
  <xsl:call-template name="draw-border-bottom"/>
</xsl:if>

The first gives the correct result, but the second paints borders only on
all but the last rowgroup. Shouldn't "... and ..." always be the same as
"not(not(...) and not(...))"? If not, is it a better way to write the above?

Gustaf


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