doubt on draft-ietf-ipsec-nat-t-ike-01.txt

["Lokesh" <lokeshnb@intotoinc.com>]

Hi All,
Following is a doubt regrading ID draft-ietf-ipsec-nat-t-ike-01
It proposes, that sending party should calculate NAT-D paylods for
both Destination and source(its own) IP address and port pairs.
that is HASH = HASH( CKY-I | CKY-R | IP | Port)
So in Normal case,(host is not multihomed) there  will be
 two NAT-D  payloads.
I want to know why it is proposed to send 2 NAT-D payloads?
For me , it looks that, there is not need for First NAT-D payload
which is Hash on Destination IP and port.
because Destination IP and Ports are not going to change in NAT,
only Source ip and source ports are changed. sending party can send
only Second NAT-D payload (HASH on its own IP and src port) ,
and receiving can determine occurance of NAT as follows.
take src ip and src port selectors from incoming packet,
prepare HASH on them and compare with HASH or NAT-D payload sent
by other peer. If match is ok, there is no NAT, if it fails, there is a NAT.

Is that Ok? or Am I   missing some point here? if so correct me please.

Thanks
-Lokesh