The End of an Era - That of The Natural Numbers
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From: Jesse F. Hughes on 4 Jun 2010 08:49 One last, pointless attempt to show Nam that his reading of Shoenfield is butt-wrong. It will not work. But here we go anyway. Let's apply the definitions found on p.19 to ~(Ex)(x = x) in the case where |M| = {}. That is not really a structure in Shoenfield's definition (a point which *still* eludes you, though it is explicit on p. 18), but no matter. We must, of course, assume that our language L has no constants and hence no closed terms. M(~(Ex)(x = x)) = T iff M((Ex)(x = x)) = F. Now, I know you don't realize Shoenfield says this, but he does. He says M(~B) = H_~(M(B)), but H_~(T) = F and H_~(F) = T. That's defined on p. 12. The *same* definition applies on p. 19. So, let's check M((Ex)(x = x)). M((Ex)(x = x)) = T iff M((x = x)_x[i]) = T for some i in L(M). L(M)is the language obtained from L by adding all the names of individuals of M, while i is a name in L(M). But M is empty, so there are *no* names in L(M). Hence, M((Ex)(x = x)) = F and thus M(~(Ex)(x = x)) = T. *That's* a simple application of Shoenfield's rules. -- Jesse F. Hughes "But a 1 in base 3 represents a larger value than a 1 in base 7." -- Albert Wagner
From: Nam Nguyen on 4 Jun 2010 09:43 Daryl McCullough wrote: > Nam Nguyen says... >> Alan Smaill wrote: >> >>> You claim that your notion of model is equivalent to Shoenfield's >>> notion. Yet Shoenfield follows Tarski's truth definition: >>> the negation of a formula is true in a structure if and only if >>> the formula is false in the structure. >> You either didn't read it carefully, or did too carefully to the >> point of being pedantic and missed what he had said there. That's all. >> For example, take the condition iii he had in defining (true) model >> that I mentioned a few times > > Yes, you clearly are very confused by it, but I really don't > understand your confusion. > > What's really weird about crackpots, is that they are not content > to just have their own alternative theory. They insist that they > understand the *standard* theory better than the non-crackpots. It's a misconception that in fora like these the bad guys are _only_ crackpots. There is a _different class_ of other bad guys too! Why don't you _technically_ answer the 4 questions (Q1 - Q4) or do the analysis about the point I raised, mentioned multiple times in condition "iii"? There are cranks who would attack people's character when they couldn't respond. Are you that much _different_ from them here?
From: Nam Nguyen on 4 Jun 2010 09:55 Jesse F. Hughes wrote: > One last, pointless attempt to show Nam that his reading of Shoenfield > is butt-wrong. > > It will not work. > > But here we go anyway. > > Let's apply the definitions found on p.19 to > > ~(Ex)(x = x) > > in the case where |M| = {}. That is not really a structure in > Shoenfield's definition (a point which *still* eludes you, though it > is explicit on p. 18), but no matter. We must, of course, assume that > our language L has no constants and hence no closed terms. > > M(~(Ex)(x = x)) = T iff M((Ex)(x = x)) = F. > > Now, I know you don't realize Shoenfield says this, but he does. He > says M(~B) = H_~(M(B)), but > > H_~(T) = F and > H_~(F) = T. > > That's defined on p. 12. The *same* definition applies on p. 19. > > So, let's check M((Ex)(x = x)). > > M((Ex)(x = x)) = T iff M((x = x)_x[i]) = T for some i in L(M). > > L(M)is the language obtained from L by adding all the names of > individuals of M, while i is a name in L(M). But M is empty, so there > are *no* names in L(M). Hence, > > M((Ex)(x = x)) = F > > and thus > > M(~(Ex)(x = x)) = T. > > *That's* a simple application of Shoenfield's rules. How is all this relevant when the case I've been talking about is the the degenerated case where U is empty? Where is your U above? Did you apply the concepts of set-hood or set membership? In my exhibits E1 and E2 before and even in the whole thread, have I ever said there's only 1 context to map a formula to the set of binary values? I really think your guys have a serious problem of hearing what people technically _did or did not_ say, hypothesize, and conclude! (So to speak, you guys are "barking at the wrong tree" and still aren't aware of the it!).
From: Nam Nguyen on 4 Jun 2010 10:18 Nam Nguyen wrote: > Jesse F. Hughes wrote: >> One last, pointless attempt to show Nam that his reading of Shoenfield >> is butt-wrong. >> It will not work. >> >> But here we go anyway. >> >> Let's apply the definitions found on p.19 to ~(Ex)(x = x) >> >> in the case where |M| = {}. That is not really a structure in >> Shoenfield's definition (a point which *still* eludes you, though it >> is explicit on p. 18), but no matter. We must, of course, assume that >> our language L has no constants and hence no closed terms. >> >> M(~(Ex)(x = x)) = T iff M((Ex)(x = x)) = F. >> >> Now, I know you don't realize Shoenfield says this, but he does. He >> says M(~B) = H_~(M(B)), but >> >> H_~(T) = F and >> H_~(F) = T. >> >> That's defined on p. 12. The *same* definition applies on p. 19. >> >> So, let's check M((Ex)(x = x)). >> >> M((Ex)(x = x)) = T iff M((x = x)_x[i]) = T for some i in L(M). >> >> L(M)is the language obtained from L by adding all the names of >> individuals of M, while i is a name in L(M). But M is empty, so there >> are *no* names in L(M). Hence, >> >> M((Ex)(x = x)) = F >> >> and thus >> >> M(~(Ex)(x = x)) = T. >> >> *That's* a simple application of Shoenfield's rules. > > How is all this relevant when the case I've been talking about > is the the degenerated case where U is empty? Where is your U above? > Did you apply the concepts of set-hood or set membership? In my > exhibits E1 and E2 before and even in the whole thread, have I > ever said there's only 1 context to map a formula to the set of > binary values? > > I really think your guys have a serious problem of hearing what > people technically _did or did not_ say, hypothesize, and conclude! > (So to speak, you guys are "barking at the wrong tree" and still > aren't aware of the it!). Look it isn't that difficult. My argument is: H implies C what you guys are "arguing" is: H' implies C Attack my H but don't attack my C with H' ! Do you understand?
From: Aatu Koskensilta on 4 Jun 2010 10:24
Nam Nguyen <namducnguyen(a)shaw.ca> writes: > Look it isn't that difficult. My argument is: > > H implies C > > what you guys are "arguing" is: > > H' implies C > > Attack my H but don't attack my C with H' ! What's at issue is not any argument you've presented, but rather your baffling and bizarre claim that every formula is true in a model with empty domain. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |