Does a proof of $0 + 0 = 0$ and $1 + (-1) = 0$ exist? The axiom of the additive neutral establishes that for every real number there is another that fulfills that when it is added the number does not change.
But the axiom does not say that for every real number the sum of that number with zero is the number.
The second I can prove from the first if I find a proof of $0 + 0 = 0$
Can someone help me?
My axioms are
$0 \in \mathbb{R}$
$1 \in \mathbb{R}$
For every $A$ and $B$  in $\mathbb{R}$
$A + B = B + A$
A + (B + C) = (A + B) + C
AB = BA
A(BC) = (AB)C
A(B + C) = AB + AC
For every A exists B such that
A + B = A
For every A exist B such thar
A + B = 0
For every A different from 0 exist B such that
AB=A
AB=1
 A: The field axioms include:

*

*There is an additive identity.

*There is a multiplicative identity.

*Every $x$ has an additive inverse.

*Every $x\neq0$ has a multiplicative inverse.

Now, suppose that the numbers $e_1$ and $e_2$ are both additive identities. Then, $e_1=e_1+e_2$, as $e_2$ is an additive identity; but $e_1+e_2=e_2$, as $e_1$ is an additive identity. Hence, $e_1=e_2$. This shows that the identity element is unique, meaning that we can speak of the additive identity. From here, it is just a matter of convention that we use the symbol "$0$" to denote the additive identity of a field. And then it follows trivially that $0+0=0$.
Now, it is common to define the real numbers axiomatically as a field satisfying certain properties, namely that it is ordered, and that it satisfies some version of completeness (e.g. the least upper bound property). Your list of axioms aren't enough. You have asserted that $0\in\mathbf R$ and that for every $A$ there is a $B$ such that $A+B=0$. However, these properties are true for every real number. You need to define $0$ by a property that marks it out as unique. And this property is typically that it is the additive identity.
Almost exactly the same remarks apply to the multiplicative identity, and additive inverses and multiplicative inverses: after showing that these are unique, we denote them by $1$, $-x$, and $x^{-1}$ (or $1/x$).
