"Dik T. Winter" wrote in message ...
In article . com "Schoenfeld" writes:
Dik T. Winter wrote:
...
Axiom: Additive Identity
for all x there exists y such that x + y = x
guarantees an additive identity.

Nope. The axiom is "there is an y such that for all x: x + y = x".
Something slightly different.

Is that your way of dodging an error you made but tacticly did not
explicitly write?

The statement,
Axiom: Additive Identity
"for all x there exists y such that x + y = x"
defines an additive identity for all x.

No it does not. Consider the following addition table:
+ a b c
a a c b
b c b a
c b a c

For each 'x' there is an 'y' such that 'x + y = x'. But there is not
an additive identity.

Your error was assuming that no such y existed because no such number
exists, but to get numbers you need a ring and the axiom above is not
compatable with that of a ring.

I am assuming nothing about numbers at all. I only look at the axiom
and sees that it does not define an additive identity.

Of course not :-)
The statement
There is a y such that for all x: P(x,y)
merely implies the statement
For all x, there is a y such that P(x,y)
but not the other way around.
Finding a counterexample is always trivial.

But that is not at stake here.
You are now arguing with Schoenie about the definition
of "additive identity" because that is the only way he's got
left to save that little bit of face he thinks he surely must
have got left - somewhere down there ;-)

Dirk Vdm