Quote from this doc page
> The $replaceWith is an alias for $replaceRoot.
This is actually not true, since the syntax for $replaceRoot is:
whereas the syntax for $replaceWith is :
I understand why it is named "alias", just I had already some developers taking this literally and failing since they did not add the newRoot part... I like to suggest to point out that this is an alias in terms of the functionality but not in terms of syntax.