[extropy-chat] Non-classic logics

[Technotranscendence]

On Monday, March 28, 2005 3:07 AM Ian Goddard iamgoddard at yahoo.com
wrote:
>> In algebra (or algebra of operators)
>> there is something called non-distributivity:
>> (AxBxC) =/= Ax(BxC) =/= (AxB)xC.
>> (Non-distributivity is different from
>> non-commutativity).
>
> I'm not sure what you mean. If by 'x' you mean the
> multiplication operator, then what you're showing is
> not true. Using '*' for multiplication, it is the case
> that
>
> (a*b*c) = a*(b*c) = (a*b)*c
>
> because multiplication is associative. It's also
> commutative
>
> a*b*c = b*c*a = c*a*b

Try that with matrices in general and you'll find they're
non-cummutative.

Also, one can set up arbitrary algebras where any of these rules fail in
the general case.  That's why there are algebras that're not
commutative -- with matrix algebra as a special case -- and ones that
are not associative -- such as Cayley algebras.

>> Is there a non-standard logic reproducing,
>> somehow, a property like this?
>
> Some classic-logic operators such as AND (&) and OR
> (v) like multiplication are also commutative and
> associative such that (where '::' indicates that the
> lefthand statement can be replaced without a loss of
> meaning by the righthand statement and visa versa):
>
> P & Q & R  ::  Q & R & P
>
> P v Q v R  ::  R v P v Q
>
>
> P & (Q & R)  ::  (P & Q) & R
>
> P v (Q v R)  ::  (P v Q) v R
>
>
> There are also distributive properties:
>
> P & (Q v R)  ::  (P & Q) v (P & R)
>
> P v (Q & R)  ::  (P v Q) & (P v R)
>
> and for the IF-THEN operator '->' here:
>
> P -> (Q v R)  ::  (P -> Q) v (P -> R)
>
> P -> (Q & R)  ::  (P -> Q) & (P -> R)

Modal operators might be an example where commutativity fails.  L~P is
not the same as ~LP -- where L stands for the "It is necessarily the
case that..."  Ditto for M~P and ~MP -- where M stands for "It is
possibly the case that..."

With classical propositional logic (cpl), wouldn't the same be true for
~ and distribution?  ~(P & Q) is not the same as (~P) & (~Q) -- think of
the case where the truth value of P does not equal the truth value of Q,
assuming bivalent logic.  In that case, the former statement is true
while the latter is false.

Regards,

Dan
    See "Free Market Anarchism: A Justification" at:
http://uweb1.superlink.net/~neptune/AnarchismJustified.html