FOM: basis vs signature; unity of human knowledge; algebra vs f.o.m.
Stephen G Simpson
simpson at math.psu.edu
Sat Mar 14 21:05:07 EST 1998
It occurs to me that Vaughan Pratt may not be grasping the distinction
between "signature" and "basis". In his posting of 10 Mar 1998
11:31:48 Vaughan writes:
> the Boolean algebra operations (those composable from whatever
> basis you choose to start from) are identical to the Boolean ring
> operations.
This is wrong! In the Boolean algebra context, it's incorrect to say
"whatever basis you choose to start from." The choice of a basis is
not up to us. The definition of Boolean algebra *specifies* a
*specific, particular, canonical* basis, namely {^,v,~}. (To be
completely correct, one could say that the signature of Boolean
algebras specifies a basis for Boolean operations under composition.)
The phrase "Boolean algebra operations" is ambiguous, but the most
correct reading is {^,v,~}, not "all operations generated from {^,v,~}
by composition".
For a while I thought Vaughan understood this, because in his posting
of 10 Mar 1998 20:39:59 he quoted Sikorski's definition of a Boolean
algebra, which explicitly specifies this particular signature. But
now it occurs to me that Vaughan may be interpreting Sikorski in an
incorrect way, viz. as saying that a Boolean algebra consists of a set
together with certain operations satisfying certain identities A1-A5
*together with all of the infinitely many operations obtained from
them by composition*! Under this incorrect reading, {^,v,~} play no
special role in the theory! Could this incorrect reading of Sikorski
be the source of our quandary? Could this be why Vaughan says:
> In the respects of language and theory therefore, Boolean algebras
> are not just isomorphic to Boolean rings, they *are* Boolean rings.
Vaughan, I now think that it's particularly pressing for you to
clarify whether you accept the notion of "signature". This notion is
defined in universal algebra books (it may also be called similarity
type). For instance, Boolean algebras have a specific signature. Do
you agree? The signature of Boolean algebras is {^,v,~}. Do you
agree? The signature of Boolean algebras is finite. Do you agree?
After we get over this hurdle, I want to explain that there are
important f.o.m. reasons for Boolean algebras to have the particular
signature {^,v,~}, rather than {+,-,.,0,1} or some other. It has to
do with the correlation of {^,v,~} with standard logical operations
{and,or,not} and set-theoretic operations
{intersection,union,complementation}. This kind of logic and set
theory is used throughout science, even in contexts where ring
operations (plus, times, etc.) are irrelevant. A key historical
reference is Boole's treatise "The Laws of Thought".
> The distinction is at best "psychological" as Sol Feferman allowed
> in a November posting.
Sol didn't say "at best psychological". He said "psychological".
(See Sol's posting of 19 Nov 1997 23:46:02). In my view, the
distinction is not only psychologically important, but also
scientifically important.
To ignore distinctions of this kind is to ignore crucial
f.o.m. issues. This is an aspect of what I have called the unity of
human knowledge. Vaughan scoffs at that phrase (see Vaughan's posting
of Tue, 21 Oct 1997 02:13:01) but believe me, it's important for
f.o.m.
It would be a gross mistake to think that algebra = logic = f.o.m.
I'm not accusing Vaughan of making this mistake, but I think he's on
the verge of it.
-- Steve
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