Saxon

You can create a function, even inline with e.g. function($seq) { $seq/. }, that will return the nodes passed in in document order. I don't know whether there is an extension function in Saxon to check whether a sequence of nodes is already in document order.

You could implement one with e.g. function($seq) { deep-equal($seq, $seq/.) }.

Comment by Michael Kay: this could give a spurious true result, since two different nodes can be deep-equal to each other.

If I read the spec correctly

(//A, //B) | ()

should return them in document order, unless some (incorrect) optimisation has elminated the empty sequence. See https://www.w3.org/TR/xpath-31/#combining_seq

You can sort any node sequence into document order (removing duplicates at the same time) using the expression

$input|()

or

$input/.

or

./$input

To check that a sequence is already in document order, with no duplicates, you could do

for-each-pair($s, tail($s), function($x, $y){ $x << $y })

and test that the result is all true, something you can do using the proposed 4.0 function fn:all(), or using every $x in ... satisfies $x.

All the solutions work. Thank you for the (very) quick response.