Optimize XPath step

If a predicate of xpath is position-independent, we don't need to create nodesets that has many duplicated nodes with different positions.

Author: tompng

lib/rexml/parsers/xpathparser.rb

@@ -655,6 +655,10 @@ def PrimaryExpr path, parsed
           n = []
           OrExpr( contents, n )
           parsed.concat(n)
+          # For xpath like `(/path[predicate1][predicate2])[predicate3][predicate4]`,
+          # add a separator mark to distinguish predicates of the inner parentheses and the outer parentheses.
+          type = n[0]
+          parsed.push(:self, :node) if type == :document || type == :child || type == :union
         end
         path
       end