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Re: Russell's Anti-Paradox -- for translating to Lojban



Yea, I'm pretty sure it's obvious to just about any Lojban user that you don't need to do proofs for Lojbanic BRIdi. Lojbanic BRIdi are statements to be understood not proven.

Wheras if we were believers in Russell's Paradox we would have to prove that our statements can have logical calculations done on them until they become Zermelo–Fraenkel axioms.

Have you ever done that or found a need to?

On 3/19/07, Andrii (lOkadin) Zvorygin <andrii.z@gmail.com > wrote:
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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