From mboxrd@z Thu Jan 1 00:00:00 1970 X-Spam-Checker-Version: SpamAssassin 3.4.4 (2020-01-24) on polar.synack.me X-Spam-Level: X-Spam-Status: No, score=-1.9 required=5.0 tests=BAYES_00 autolearn=ham autolearn_force=no version=3.4.4 X-Google-Thread: a07f3367d7,39579ad87542da0e X-Google-Attributes: gida07f3367d7,public,usenet X-Google-NewGroupId: yes X-Google-Language: ENGLISH,ASCII-7-bit X-Received: by 10.180.11.232 with SMTP id t8mr726467wib.3.1369279638206; Wed, 22 May 2013 20:27:18 -0700 (PDT) Path: fw11ni1151wic.0!nntp.google.com!feeder1.cambriumusenet.nl!feed.tweaknews.nl!193.141.40.65.MISMATCH!npeer.de.kpn-eurorings.net!npeer-ng0.de.kpn-eurorings.net!border2.nntp.ams2.giganews.com!border4.nntp.ams.giganews.com!border2.nntp.ams.giganews.com!nntp.giganews.com!news.panservice.it!fu-berlin.de!uni-berlin.de!individual.net!not-for-mail From: Niklas Holsti Newsgroups: comp.lang.ada Subject: Re: Seeking for papers about tagged types vs access to subprograms Date: Fri, 17 May 2013 08:25:37 +0300 Organization: Tidorum Ltd Message-ID: References: <1oy5rmprgawqs.1jz36okze0xju$.dlg@40tude.net> <1q2ql1e4rcgko.diszzq1mhaq8$.dlg@40tude.net> <518dedd4$0$6581$9b4e6d93@newsspool3.arcor-online.net> <1um7tijeo609b$.1gtdijp0acfmn$.dlg@40tude.net> <1nkyb845dehcu.1sd90udwsrpdu.dlg@40tude.net> <1mg9eepp12ood$.14lj7s8a7eygd$.dlg@40tude.net> <1cj8b5amtua30.1foe0rdpldt2.dlg@40tude.net> <5195144f$0$6558$9b4e6d93@newsspool4.arcor-online.net> Mime-Version: 1.0 X-Trace: individual.net aNGKFigPsZNxj+5b1H6Mdg9din5LsZ7H//hvK+4KOaVg5bJt2f Cancel-Lock: sha1:14w7EY7TRQjssiZ332rIr3wHisE= User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.6; rv:17.0) Gecko/20130328 Thunderbird/17.0.5 In-Reply-To: Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit Date: 2013-05-17T08:25:37+03:00 List-Id: On 13-05-17 02:20 , Peter C. Chapin wrote: > On Fri, 17 May 2013, Niklas Holsti wrote: > >> A proof seems simple enough, assuming that the set of inputs is >> infinite and enumerable: if you want to decide whether f1 and f2 are >> equivalent, make a program P that iteratively generates (enumerates) >> all possible inputs x; computes f1(x) and f2(x); and stops if f1(x) /= >> f2(x). This program P halts if and only if f1 /= f2. Since halting is >> undecidable, so therefore is function equality. > > You've shown that the method of enumerating all possible argument values > and testing f1(x) = f2(x) doesn't work. Yes, my argument is wrong (the reduction should go the other way). I tried to cancel my post, but too late.... sorry. I should not try to write proofs of undecidability at one hour past midnight. -- Niklas Holsti Tidorum Ltd niklas holsti tidorum fi . @ .