On the Implication of Addition Axioms

[OhMyMarkov]

Hello everyone!

I was trying to prove the propositions that follow the addition axioms as a revision, I got a different proof for the following proposition:

If $x+y=x+z$ then $y=z$

My proof was the following:
$x+y=x+z$, $(-x)+x+y=(-x)+x+z$, $0+y=0+z$, $y=z$

Rudin however, in his book, provides a different proof:
$y=0+y=(-x+x)+y=-x+(x+y)=-x+(x+z)=(-x+x)+z=0+z=z$

This is how I was thinking, I start from the condition, to reach the result. Is my proof incorrect?
Thanks!

 

[Ackbach]

I think both proofs are fine, so long as you can assume the following:

1. Existence and uniqueness of additive inverses
2. Existence and uniqueness of the additive identity
3. Associativity of addition

 

[Evgeny.Makarov]

It is not necessary to assume the uniqueness of additive inverses and the additive identity.

I agree that both proofs are correct and essentially the same. In some formal calculi, which identify proofs with simple variations, they may literally be the same proof.

 

[Ackbach]

It is not necessary to assume the uniqueness of additive inverses and the additive identity.

I agree that both proofs are correct and essentially the same. In some formal calculi, which identify proofs with simple variations, they may literally be the same proof.

Are you sure you don't need uniqueness of, say, the identity? I'm just wondering how
$$y=0+y=(-x+x)+y$$
would work if you didn't have uniqueness. How could you be sure that $-x+x=0$ and not $-x+x=0'$?

 

[Evgeny.Makarov]

The OP's proof does not use the uniqueness.

Are you sure you don't need uniqueness of, say, the identity? I'm just wondering how
$$y=0+y=(-x+x)+y$$
would work if you didn't have uniqueness. How could you be sure that $-x+x=0$ and not $-x+x=0'$?

You could use the 0 for which -x + x = 0. Then you rewrite y as 0 + y using that particular 0.