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   RE: [xml-dev] Quick question regarding XPath

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# is it possible to say in XPath
# a//@b
# is the above equivalent to:
# (a/@b | a//*/@b)

Yes. I think the easiest proof of this (using constructs that are allowed in
XPath 2.0 along the way) is




which means

a/(descendant::node | self::node())/@b

which means

a/descendant::node()/@b | a/self::node()/@b

which means

a/child::node()/descendant-or-self::node()/@b | a/@b

which means

a/descendant-or-self::node()/child::node()/@b | a/@b

which means (given that nodes other than elements have no attributes)

a//*/@b | a/@b

Of course this assumes some equivalences which I haven't tried to prove
formally, for example that "|" distributes over "/". 

Michael Kay


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