proof of one-wayness and P != NP

From: Bartosz Zoltak (X_at_vmpcfunction.com;)
Date: 10/18/05 

Date: Tue, 18 Oct 2005 10:50:07 +0200

Proving one-wayness of a function would show P != NP. But what if the
set of values has fewer (say 5-fold) elements than the set of
arguments and the function can produce equal values from different
arguments?

Then the proof would show that finding ANY of the arguments which
produce a given value is hard on the average case - would such proof
still be showing that P != NP?

Thanks 

--
Bartosz Zoltak
http://www.vmpcfunction.com
X@vmpcfunction.com; X=bzoltak
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