The End of an Era - That of The Natural Numbers
in [Math]
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From: William Hughes on 19 May 2010 01:51 On May 19, 2:26 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > In the current FOL, > > "F is valid in every model of T" > > is equivalent to > > "T is consistent _and_ F is provable". Nope, in the current FOL, F is valid in every model of T does not imply that T has at least one model. - William Hughes
From: Nam Nguyen on 19 May 2010 02:06 William Hughes wrote: > On May 19, 2:26 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > >> In the current FOL, >> >> "F is valid in every model of T" >> >> is equivalent to >> >> "T is consistent _and_ F is provable". > > > Nope, in the current FOL, > F is valid in every model of T > does not imply that T has at least one model. Let me ask you this: as a _fact_ , does the T = {(x=x) /\ ~(x=x)} have any (true) model? And what does the implication you've just mentioned have to do with this fact?
From: William Hughes on 19 May 2010 02:22 On May 19, 3:06 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > William Hughes wrote: > Let me ask you this: as a _fact_ , does the T = {(x=x) /\ ~(x=x)} > have any (true) model? And what does the implication you've just > mentioned have to do with this fact? In the current FOL F is valid in every model of {(x=x) /\ ~(x=x)} (*) does not imply {(x=x) /\ ~(x=x)} has a model. Therefore, you cannot use the fact that {(x=x) /\ ~(x=x)} does not have a model to show (*) is false. Indeed, in the current FOL, you can only show that (*) is false by finding a model for {(x=x) /\ ~(x=x)} in which F is not valid. - William Hughes
From: Nam Nguyen on 19 May 2010 02:27 William Hughes wrote: > On May 19, 3:06 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> William Hughes wrote: > >> Let me ask you this: as a _fact_ , does the T = {(x=x) /\ ~(x=x)} >> have any (true) model? And what does the implication you've just >> mentioned have to do with this fact? > > In the current FOL > > F is valid in every model of {(x=x) /\ ~(x=x)} (*) > > does not imply {(x=x) /\ ~(x=x)} has a model. Therefore, > you cannot use the fact that {(x=x) /\ ~(x=x)} does not > have a model to show (*) is false. Indeed, in the current > FOL, you can only show that (*) is false > by finding a model for {(x=x) /\ ~(x=x)} > in which F is not valid. What's your definition of "valid" in this context is?
From: William Hughes on 19 May 2010 02:47
On May 19, 3:27 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > William Hughes wrote: > > In the current FOL > > > F is valid in every model of {(x=x) /\ ~(x=x)} (*) > > > does not imply {(x=x) /\ ~(x=x)} has a model. Therefore, > > you cannot use the fact that {(x=x) /\ ~(x=x)} does not > > have a model to show (*) is false. Indeed, in the current > > FOL, you can only show that (*) is false > > by finding a model for {(x=x) /\ ~(x=x)} > > in which F is not valid. > > What's your definition of "valid" in this context is? "valid" = "true", So Indeed, in the current FOL you can only show that (*) is false, by finding a model for {(x=x) /\ ~(x=x)} in which F is false. - William Hughes |