[FOM] Why inclusive disjunction?
A. Mani
a_mani_sc_gs at yahoo.co.in
Thu Jan 11 17:19:47 EST 2007
On Wednesday 10 Jan 2007 13:34, Lucius Schoenbaum wrote:
> It might have something to do with the early discovery of the Boolean
> structure and the analogy with set union and intersection. Boole in
> an investigation of the laws of thought already has addition for
> union, concatenative multiplication for intersection, and gives a
> distributive law x(y+z)=xy+xz.
Boole did not exactly inspire the boolean algebra. His partial algebraic
semantics (badly formulated) for the laws of thought was motivated by a
desire to force the laws of thought to follow the rules of arithmetic.
Schroder, Pierce and others can be credited with preferring the inclusive
\vee.
The principled reason is in the difficulties in dealing with concepts of the
dual in formal logic. Philosophers write quite a bit on the associated
functors.
A. Mani
Member, Cal. Math. Soc
Send instant messages to your online friends http://in.messenger.yahoo.com
More information about the FOM
mailing list