From nobody@digitalkingdom.org Mon Mar 19 14:43:36 2007 Received: with ECARTIS (v1.0.0; list lojban-list); Mon, 19 Mar 2007 14:43:37 -0700 (PDT) Received: from nobody by chain.digitalkingdom.org with local (Exim 4.63) (envelope-from ) id 1HTPdM-00055o-0m for lojban-list-real@lojban.org; Mon, 19 Mar 2007 14:43:16 -0700 Received: from an-out-0708.google.com ([209.85.132.245]) by chain.digitalkingdom.org with esmtp (Exim 4.63) (envelope-from ) id 1HTPdC-00055Q-GC for lojban-list@lojban.org; Mon, 19 Mar 2007 14:43:13 -0700 Received: by an-out-0708.google.com with SMTP id b8so1255888ana for ; Mon, 19 Mar 2007 14:43:04 -0700 (PDT) DKIM-Signature: a=rsa-sha1; c=relaxed/relaxed; d=gmail.com; s=beta; h=domainkey-signature:received:received:message-id:date:from:to:subject:in-reply-to:mime-version:content-type:references; b=hNk7CR9llBnoHx2HL6g3Car1h0eWW9gMFLEs5XBxDQm/AJ/oNGQeIgMQlbzg2UNXiYpG5wZMJoubOna+DV6GcqKR6z8+bylN4IxDBKvjEJ83w9dTVSYF98gZxEHGzorG6WJfUt5onJF9eFfH3lrKBJigN/ke/QFcBf3uR2MHXB4= DomainKey-Signature: a=rsa-sha1; c=nofws; d=gmail.com; s=beta; h=received:message-id:date:from:to:subject:in-reply-to:mime-version:content-type:references; b=YpxVVCaXrDpy/Wl+vqGA+YafRcViHOKi9voYuoUrf/0vqMtYztAiSQfqfa1sLBKxrJDjjfVuFQOfyQcTw5Ujtc3xV65n8asryoporzGVgFAEsP5nmkq/SA0Hx9x6P57YbuSPVp/d0HeP2Bci00GfFkm4hgjnrJLfqdKayZKpSjY= Received: by 10.100.195.10 with SMTP id s10mr4218364anf.1174340584401; Mon, 19 Mar 2007 14:43:04 -0700 (PDT) Received: by 10.100.43.15 with HTTP; Mon, 19 Mar 2007 14:43:04 -0700 (PDT) Message-ID: Date: Mon, 19 Mar 2007 17:43:04 -0400 From: "Andrii (lOkadin) Zvorygin" To: "lojban-list@lojban.org" Subject: [lojban] Fwd: Russell's Anti-Paradox -- for translating to Lojban In-Reply-To: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_Part_116027_1613769.1174340584351" References: X-Spam-Score: -2.0 X-Spam-Score-Int: -19 X-Spam-Bar: -- X-archive-position: 13617 X-ecartis-version: Ecartis v1.0.0 Sender: lojban-list-bounce@lojban.org Errors-to: lojban-list-bounce@lojban.org X-original-sender: andrii.z@gmail.com Precedence: bulk Reply-to: lojban-list@lojban.org X-list: lojban-list ------=_Part_116027_1613769.1174340584351 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 7bit Content-Disposition: inline k i did some lojban translation. you have the original to reference if something doesn't make sense. The Russell's paradox has been around a long time. I really wonder why no one has noticed that "barber shop" example is a false analogy to the Universal Set. I won't get into the barbershop example as it would only distract from the logic but I do have a full thread da na CMIma da .i ganai da CMIma da gi da du da It is actually inconcievable that da is an Element of da. It is concievable that da is da. Though the contradiction is also true for a complete system it would still just be a reference to the same object -- sa'u da CMIma da du da. ni'o ny du ro da poi RARna NAMcu .i la'i .ny. du lu'i ro RARna'u gi 1. now we have a set la'iby. du lu'i ny. ce vei ny. ve'o 2. ge lu'ida CMIma la'iby. gi ta'o lu'ida nadu la'iby. .i ma'a JIMpe ledu'u da de DRAta zoi.gy. have a different number of elements as well as transfinite cardinality .gy. ta zoi .gy. Note the { } are necessary to add the piece of information distinguishing this as a set rather than an arbitrary list of unrelated numbers.gy. 3. ge daduda giku'i da naCMIma da ni'o ganai la'i da du lu'i da ce de gi daCMImada .i ku'i ge la'i da nadu lu'i da gi da nadu da .ui.u'i Let f(x) be any formula of first order logic in which da is a free variable . *Definition*. The *collection* la'i abu, denoted lu'i da zo'u da CMIma *la'i abu .o *f(x) , is the individual la'i abu satisfying lu'i roda zo'u da CMIma la'i .abu. .o f(x) to which Russel says: ni'o roda zo'u da CMIma da .o da naCMIma da .idu da CMIma da .o da naCMIma da to which one might say: da du da .o da naCMIma da da .o da naCMIma da da naCMIma da is actually a fundamental truth. As is da CMIma da du da So da naCMIma da can be subsititutied with JETnu da .o JETnu da AKA a contradiction So as we already know that a contradiction can prove any concievable thing. simple proof: roda can not be proved from a contradiction. roda can be proved from a contradiction. Can't argue with that and it's not a circular argument. Multiplication begins! as now we have more than a single item. so wheras with a circle you have da and da, wheras with a contradiction you have da .i nada, which is something that is not da, so something else! Do you understand? Now we can increase the amount of things by adding as many nada's as we desire and renaming them de .a di .a LERfubu .a roda :D there you have it complete mathematical theory. 1/0 -- It should be noted that this email was generated by Your subconciousness. You should note that the email you get is the email you expect -- unless you expect an unexpected email. Have a nice day! Be Happy! :D join la.ma'aSELTcan. for more information. -- It should be noted that this email was generated by Your subconciousness. You should note that the email you get is the email you expect -- unless you expect an unexpected email. Have a nice day! Be Happy! :D join la.ma'aSELTcan. for more information. -- It should be noted that this email was generated by Your subconciousness. You should note that the email you get is the email you expect -- unless you expect an unexpected email. Have a nice day! Be Happy! :D join la.ma'aSELTcan. for more information. -- It should be noted that this email was generated by Your subconciousness. You should note that the email you get is the email you expect -- unless you expect an unexpected email. Have a nice day! Be Happy! :D join la.ma'aSELTcan. for more information. ------=_Part_116027_1613769.1174340584351 Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: 7bit Content-Disposition: inline k i did some lojban translation. you have the original to reference if something doesn't make sense.


The Russell's paradox has been around a long time. I really wonder why no one has noticed  that "barber shop" example is a false analogy to the Universal Set. I won't get into the barbershop example as it would only distract from the logic but I do have a full thread

da  na CMIma da .i ganai da CMIma da gi da du da

It is actually inconcievable that da is an Element of da. It is concievable that da is da. Though the contradiction is also true for a complete system it would still just be a reference to the same object -- sa'u da CMIma da  du da.

ni'o ny du ro da poi RARna NAMcu
.i la'i .ny. du lu'i ro RARna'u gi

  1. now we have a set la'iby. du lu'i ny. ce vei ny. ve'o
  2. ge lu'ida CMIma la'iby.  gi ta'o lu'ida  nadu la'iby.  .i ma'a JIMpe ledu'u da de DRAta zoi.gy. have a different number of elements as well as transfinite cardinality .gy. ta zoi .gy. Note the { } are necessary to add the piece of information distinguishing this as a set rather than an arbitrary list of unrelated numbers.gy.
  3. ge daduda giku'i da naCMIma da


ni'o ganai la'i da du lu'i da ce de  gi daCMImada
.i   ku'i ge la'i da nadu lu'i da gi da nadu da  .ui.u'i

Let f(x) be any formula of first order logic in which da is a free variable.

Definition. The collection la'i abu, denoted lu'i da  zo'u da CMIma la'i abu .o f(x) , is the individual la'i abu satisfying  lu'i roda  zo'u da CMIma la'i .abu. .o f(x)



to which Russel says: 
ni'o roda zo'u da CMIma da .o da naCMIma da
.idu da CMIma da .o da naCMIma da

to which one might say:

da du da .o da naCMIma da
da .o da naCMIma da

da naCMIma da is actually a fundamental truth. As is da CMIma da du da
So da naCMIma da can be subsititutied with JETnu

da .o JETnu

da                                  AKA a contradiction

So as we already know that a contradiction can prove any concievable thing.

simple proof:  roda can not be proved from a contradiction. roda can be proved from a contradiction.

Can't argue with that and it's not a circular argument.  Multiplication begins! as now we have more than a single item. so wheras with a circle you have da and da, wheras with a contradiction you have da .i nada, which is something that is not da, so something else! Do you understand? Now we can increase the amount of things by adding as many nada's as we desire and renaming them de .a di .a LERfubu .a roda :D there you have it complete mathematical theory.


1/0

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