sci.logic
[Top] [All Lists]

Re: The L3 Revolution

Subject: Re: The L3 Revolution
From: "Confutus"
Date: 27 Oct 2006 18:25:44 -0700
Newsgroups: sci.logic
Jan Burse wrote:
> http://www.sapiencekb.com/logic3rev.htm
>
> Cite:
>    "Negation and the non-strict conditional are defined by these axioms.
>     Other connectives can be defined. The first four are standard. The
>     next two have apparently not been previously defined and used in
>     Lukasiewicz logic: an oversight which has prevented this logic from
>     being properly appreciated for nearly 90 years. The last two have
>     been previously defined but little used."
>
>    can be defined & have not been previously defined
>        --> logic not properly appreciated
>
> I don't believe the above claim.

I can hardly blame you, since  I can scaredly believe it myself. Hasn't
*anyone* put all this together before?  Where's the catch?  But
disbelief isn't disproof.

> []_ and <>_ are already used elsewhere,
> namely in _+ and _- as follows:
>
>     A+ == []A
>     A- == ~<>A

Well, yes, and the same functions appear in Lukasiewicz original
treatments also, I do believe. But <>_ and {}_  aren't the ones  I was
referring to. It's [](_ -> _}  and  [](_< -> _} applied to the
Lukasiewicz conditional that seem to have gone unappreciated.

> This can be used to translate L3 formulas
> over 3-valued propositional variables V
> into two boolean formulas over 2-valued
> propositional variables V+ and V-, with the
> following rules:
>
>    (A v B)+ := A+ v B+
>    (A & B)+ := A+ & B+
>    (~A)+ := A-
>    (P)+ := P+
>    (A v B)- := A- & B-
>    (A & B)- := A- v B-
>    (~A)- := A+
>    (P)- := P-
>
> Example:
>
>    (~P v Q)+ = P- v Q+
>    (~P v Q)- = P+ & Q-
>
> The advantage of the _+ and _- translation
> is that one, depending on its need, can
> directly read off []_ and ~<>_ without
> going through 3 valued logic.

Granted, but without a well-behaved conditional this is only a fragment
of a full-featured logic.

> One can also reconstruct the 3 valued value,
> as the following holds:
>
>    A == A+ v ~(u v A-)
>
> Note the above needs the constant u,
> I am not sure whether this is part
> of your L3, or whether a formula can
> be constructed in L3 that gives u.

No, this isn't a part of  L3.  The truth value U would work here, but
I use the truth values only to substitute into formulas in order to
evalute them, not as constants, and  there is no formula  that yields
only the truth value U.


<Prev in Thread] Current Thread [Next in Thread>
Privacy Policy