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[xsl] Can the default function namespace be undeclared in XSLT?

2014-06-01 16:16:03
 I want to run the XSLT - analog of the last example (written in
XQuery) in section 17.5 of the F & O 3.0 Spec
(http://www.w3.org/TR/xpath-functions-30/#constructor-functions-for-user-defined-types):

"declare default function namespace ''; apple(17)"

However I cannot find any XSLT "default-function-namespace" attribute
in the XSLT 3.0 Last Call WD.

How can this be done in XSLT 3.0?

-- 
Cheers,
Dimitre Novatchev
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To avoid situations in which you might make mistakes may be the
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