Under the normalization rules specified in formal semantics, the query:

(xs:anyURI("example.com/"), 
 1, 
 QName("example.com", "ncname"), 
 false(), 
 xs:hexBinary(\"FF\")) 
  = 
(xs:anyURI("example.com/NOT\), 
 0, 
 QName("example.com", "p:ncname"), 
 true(), 
 xs:hexBinary(\"EF\"))

expands to:

some $v1 in fn:data(([Expr1]Expr)) satisfies
  some $v2 in fn:data(([Expr2]Expr)) satisfies
    let $u1 := fs:convert-operand($v1, $v2) return
      let $u2 := fs:convert-operand($v2, $v1) return
        fs:eq($u1, $u2)

Now:

"When evaluating a general comparison in which either operand is a sequence of items, an implementation may return true as soon as it finds an item in the first operand and an item in the second operand that have the required magnitude relationship. Similarly, a general comparison may raise a dynamic error as soon as it encounters an error in evaluating either operand, or in comparing a pair of items from the two operands. As a result of these rules, the result of a general comparison is not deterministic in the presence of errors."

Since fs:eq is only defined when the two arguments have the same type, the comparison may result in a type checking error, which should be reflected in the expected results of this test.

Is there an argument for the following:

some $v1 in fn:data(([Expr1]Expr)) satisfies
  some $v2 in fn:data(([Expr2]Expr)) satisfies
    let $u1 := fs:convert-operand($v1, $v2) return
      let $u2 := fs:convert-operand($v2, $v1) return
        (if fs:eq is defined on type($u1), type($u2)
         then false
         else fs:eq($u1, $u2))

?