From: "x" Newsgroups: comp.databases.theory References: <40fe3be3$1@post.usenet.com> <40feafc9@post.usenet.com> Subject: Re: thinking about UPDATE Date: Mon, 26 Jul 2004 16:07:11 +0300 Lines: 34 X-Priority: 3 X-MSMail-Priority: Normal X-Newsreader: Microsoft Outlook Express 6.00.2800.1437 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1441 NNTP-Posting-Host: 81.196.98.200 X-Original-NNTP-Posting-Host: 81.196.98.200 Message-ID: <410501fc$1@post.usenet.com> X-Trace: post.usenet.com 1090847228 81.196.98.200 (26 Jul 2004 08:07:08 -0500) Organization: Usenet.com http://www.usenet.com 100,000+ UNCENSORED Newsgroups. The #1 Usenet Service on the Planet! X-Comments: This message was posted through Usenet.com's, FREE publicly accessible Usenet Server - "post.usenet.com" X-Comments2: IMPORTANT: Usenet.com does not condone, nor support, spam or any illegal or copyrighted postings. X-Comments3: IMPORTANT: Under NO circumstances will postings containing illegal or copyrighted material through this service be tolerated!! X-Report: Please report illegal or inappropriate use to X-Abuse-Info: Please be sure to forward a copy of ALL headers, INCLUDING the body (DO NOT SEND ATTACHMENTS) Path: news.easynews.com!core-easynews!newsfeed1.easynews.com!easynews.com!easynews!news-out2.nntp.be!local!post.usenet.com!not-for-mail Xref: core-easynews comp.databases.theory:37352 X-Received-Date: Mon, 26 Jul 2004 06:06:16 MST (news.easynews.com) **** Post for FREE via your newsreader at post.usenet.com **** "Jan Hidders" wrote in message news:pan.2004.07.22.18.06.23.604113@REMOVETHIS.pandora.be... > If K is the set of original candidate keys and P is the set of attribute > on which we project then: > - Let K' = { k in K | k subset P } > - If K' is not empty then return K' otherwise { P } > > Also I note that my post here is simply thinking out the > > cases, and I have no formalism to back this up. Does anyone > > have any suggestions as to a formalism to apply to either > > prove or disprove the above? > Yep, normalization theory. You can translate the candidate keys to > functional dependencies and these always also hold for the projection, and > those that hold for the projection also hold for the original relation. So > the proof of completeness goes something like this. > Suppose that in the projection we have a candidate key A that is not in > K'. Then the FD A->P holds in the projection, but also in the original > relation. However, the only FDs that hold in the original relation are > those that are implied by its CKs so A must have already been a CK in the > original relation. But then A is by definition in K' which leads to a > contradiction. So the original assumption that the projection has a CK > that is not in K' must be false. QED What made you think this: "However, the only FDs that hold in the original relation are those that are implied by its CKs" -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= *** Usenet.com - The #1 Usenet Newsgroup Service on The Planet! *** http://www.usenet.com Unlimited Download - 19 Seperate Servers - 90,000 groups - Uncensored -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=