Is there any difference between md5 and sha1 in this situation?

It is known that

1.    if   ( md5(a)   == md5(b)   )
2.    then ( md5(a.z) == md5(b.z) )
3.    but  ( md5(z.a) != md5(z.b) )

where the dots concatenate the strings.

EDIT ---

Here you can find a and b:
http://www.mscs.dal.ca/~selinger/md5collision/

Check these links:
hexpaste.com/qzNCBRYb/1 - this is a.md5(a)."kutykurutty"
hexpaste.com/mSXMl13A/1 - this is b.md5(b)."kutykurutty"

They share the same md5 hash, yet they are different. But you can call these strings a' and b', because they have the same md5.

--- EDIT

What happens in the second row if we change all the md5 to sha1? So:

1.    if   ( sha1(c)   == sha1(d)   )
2.    then ( sha1(c.z) ?= sha1(d.z) )

I couldn't find two different strings with same sha1, that's why I'm asking this. Are there any other interesting "rules" about sha1?

Solution

SHA1 will behave exactly like MD5 in this scenario.

The only two references I have found are the following -

http://www.iaik.tugraz.at/content/research/krypto/sha1/MeaningfulCollisions.php
http://www.schneier.com/blog/archives/2005/02/sha1_broken.html#c1654 (See comment by David Schwartz)

From the IAIK website -

Note that for colliding SHA-1 message pairs (as for all other hash functions following a similar design principle) it is always possible to append suffixes to both messages as long as they are the same.

I don't think anybody has found two colliding strings for SHA1, so this is mostly an academic discussion. But from what I understand, when a collision is discovered, it should be possible to create several other collisions by using this property.