On Sun, Sep 30, 2012 at 1:36 AM, Chris Menzel <chris.menzel@xxxxxxxxx> wrote:
On Sat, Sep 29, 2012 at 8:05 PM, William Frank <williamf.frank@xxxxxxxxx> wrote:
On Sat, Sep 29, 2012 at 7:03 PM, Chris Menzel <chris.menzel@xxxxxxxxx> wrote:
On Fri, Sep 28, 2012 at 11:54 AM, John F Sowa <sowa@xxxxxxxxxxx> wrote:
I would emphasize that set theory has two operators (subset and isIn),
but mereology has only one operator (partOf).
John, virtually every text on set theory presents the axioms with just a single binary predicate "∈" for membership. The subset relation is always defined; using "isin" instead of "∈":
What does this imply to you?
It doesn't imply anything to me. It is just a simple fact about how the subset relation is introduced into set theory.
arithmetic has the operations plus and minus and times and successor. Of course, most texts define all of them from sucessor.
I
hope I would be as unlikely to be '*quite* mistaken' as if someone
would make clear assertions about operations in set theory and yet not
know that subset could be defined in terms of membership. If I knew
that little about the subject of axiomatized arithmetic, I hope I would
not speak about it.
It
is impossible to define addition in terms of successor and
multiplication in terms of addition and successor. You appear to be
mistaking the usual recursive axioms for those operators for
definitions. They are not. Full arithmetic requires all three
operations. (Interestingly, exponentiation can be defined in terms of
addition and multiplication.)
I think what you meant was, 'in a first order axiomatization of arithmetic'. Otherwise, for a start
Defintion of '+'
for every f, f = +
means (or iff)
for every x, f(x,0) = x
and
for every x, y, z if f(x,y) = z.
then s(fx,y)) = s(z).
Then,
if you don't like this form of a definition, with the right
circumlocution, it can be made to suit whatever form you want a
definition to have. I seem to recall that one just has to replace the
function with a property, and then show that the for every first two
arguments of the property there is exactly one third argument, and then
use the iota operator to say + is *the* property P such that ..... As,
long, of course, as one does not restrict oneself from quantifying over
properties. I seem to recall I read a paper at a meeting of the Association of Symbolic Logic on this once.
Interesting
and profound, but it remains the case that all three are part of
arithmetic. Not just the ones you choose in your axoimatization to be
primitive. Instead, they can all be defined from + and 1, if one
chose. There was a time before successor was discovered. When it was,
did the others go away? In propositional logic, we can use all the
natural deduction operators, and after defining the inference rules for
them all, prove many equivelences, or instead, take our pick of a pair
such as not and or, or use nor or nand only as primitives. I have
found that some people, depending on what course they happen to have
taken, believe that if a then b "really is" not a or b, etc.
I have no idea what you are talking about.
Sorry.
I read in to your reason for saying that subset can be defined in
terms of membership, in almost every axiomatization. In doing this, I
was quite mistaken, not just in misreading, but in reading in instead of just asking, about a subject I do know very little (other people's intentions).
Wm
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You are quite mistaken. It is impossible to define addition in terms of successor and multiplication in terms of addition and successor. You appear to be mistaking the usual recursive axioms for those operators for definitions. They are not. Full arithmetic requires all three operations. (Interestingly, exponentiation can be defined in terms of addition and multiplication.)
Interesting and profound, but it remains the case that all three are part of arithmetic. Not just the ones you choose in your axoimatization to be primitive. Instead, they can all be defined from + and 1, if one chose. There was a time before successor was discovered. When it was, did the others go away? In propositional logic, we can use all the natural deduction operators, and after defining the inference rules for them all, prove many equivelences, or instead, take our pick of a pair such as not and or, or use nor or nand only as primitives. I have found that some people, depending on what course they happen to have taken, believe that if a then b "really is" not a or b, etc.
I have no idea what you are talking about.
-chris
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