Re: NULLs: theoretical problems?

On Aug 10, 4:52 pm, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> "David Portas" <REMOVE_BEFORE_REPLYING_dpor..._at_acm.org> wrote innews:NM-dncYFOuhqBybb4p2dnAA@giganews.com:
>
> > "paul c" <toledobythe..._at_oohay.ac> wrote in message
> >news:JXLui.45171$rX4.26997_at_pd7urf2no...
>
> >> (even though I'm not sure in "s{X} = t{X} implies s{Y} = t{Y}"
> >> whether "implies" stands for logical implication.)
>
> > Good catch. It seems that logical implication is not well defined for
> > three-value logic.
>
> It is not that three-valued implication is not 'well defined' whatever it
> means. As a matter of fact, there are a few competing definitions to
> choose from, Lukaciewicz's, Kleene's and someone else's whose name I do
> not recall. They define implication in the usual way, with the truth
> table.

I wonder if 3-rd value logic interpretation is trivial. Take any boolean algebra that is more than 2 valued, and partition its elements into 3 equivalence classes. For example, one may define True as maximal element, False as a minimal one, and combine all the rest into Unknown. For four element BA we have:

00 -- False
01 -- Unknown
10 -- Unknown
11 -- True

Sure in this model formal implication "Unknown -> Unknown" evaluates to True or Unknown:

"01 -> 01" = "01 \/ ~01" = "01 \/ 10" = "11" -- true

on the other hand

"01 -> 10" = "01 \/ ~10" = "01 \/ 01" = "01" -- unknown

So the problem is to make the partition of BA elements to respect BA operations, so that the later can be defined consistently. Apparently, one can have consistent 4 valued logic, but not 3 valued one. Am I missing anything? Received on Sat Aug 11 2007 - 03:59:58 CEST