Re: A simple proof involving Peano's axioms

On Jun 4, 1:15 am, Bill Dubuque <w...@xxxxxxxxxxxxxxxxxxxx> wrote:
Mariano Su?rez-Alvarez <mariano.suarezalva...@xxxxxxxxx> wrote:
On Jun 3, 8:08 pm, Bill Dubuque <w...@xxxxxxxxxxxxxxxxxxxx> wrote:
Mariano Su?rez-Alvarez <mariano.suarezalva...@xxxxxxxxx> wrote:
Bill Dubuque <w...@xxxxxxxxxxxxxxxxxxxx> wrote:
Dave <dave_and_da...@xxxxxxxx> wrote:
On May 31, 3:33?pm, <pers...@xxxxxxxxxxxxxx> wrote:

1) I want to prove, -(-a) = a  [via Ring axioms -wgd]

Is it possible to prove it without resorting to induction axiom?.
Looks to me like you can't. In the particilar pdf above, the author
claims it should be possible to prove a=-(-a), without using the
induction axiom. (look at exercise 1). I wonder how that is possible.

-(-a) = (-(-a)) + 0 = 0 + (-(-a)) = (a + (-a)) + (-(-a)) = a + ((-a) +
(-(-a))) = a + 0 = a
                           ------
SIMPLER  Evaluate  -(-a) + -a + a   in two ways
                   ----------
 i.e.   -(-a), a  are both inverses for -a,  and
                      ------
Inverses are Unique:  -n + n + -n'
                           ------
                 ->   -n   =   -n'   via over/underlined terms = 0

For a deeper understanding see my many posts on the Law of Signs
http://google.com/groups/search?q=group:*math*+wgd+law+signs&scoring=d

That's only `simpler' in that you are using identities like
`(-a) + a = 0' and `0 + a = 0' which are not provided by the axioms
presented in the pdf linked to by the OP. If you refrain to do that,
then your proof is neither simpler nor shorter.

'Simpler' has many connotations in mathematics. In particular, it need
not necessarily denote purely syntactic (shorter) or semantic (elementary)
objects. Above it is intended to denote a conceptually simpler proof.
You'd've understood that if you read my many prior posts linked above.

Well, I do not find it conceptually simpler. At all. I cannot see it
as  anything else but a different spelling out of what Dave wrote.
A case of YMMV, I guess. [...]

I suspect you're missing my point. See my most recent reply to Dave.

Rather, I do not think your point is compelling enough
in this context. ;-)

-- m


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