I am slowly working my way trough "Mathematical Analysis" by Apostol, and I decided to try my hand at a few proofs. (I asked related questions here and here, so some of the work here might be from the answers there)
So far we are only given the axioms:
Axiom 1: Commutative Laws
$x+y = y + x$ , $xy=yx$
Axiom 2: Associative Laws
$x+(y+x) = (x+y)+z$ , $x(yz) = (xy)z$
Axiom 3 Distributive Law
$x(y+z)=xy+yz$
Axiom 4
Given any two real numbers x and y, there exists a real number z such that x + z = y. This z is denoted by y-x; the number x-x is denoted by 0. (It can be proved that 0 is independent of x.) We write -x for 0-x and call -x the negative of x
Proof 1: $x + 0 = x$
From axiom 4, we are guaranteed a $z$ such that $x+z = x$.
That $z$ is equal to $x-x=0$,
By substitution, $x + 0 = x$
Comments/concerns: I am not sure weather the $0$ in Axiom 4 is used just as a symbol/placeholder for $x-x$, or if it is actually meant to be the actual integer between $-1$ and $1$. That is, I am not sure if it's implied that $x-x$ evaluates to the same thing as $y-y$. I;m not sure that this is what is meant in axiom 4 by"it can be proved that $0$ is independent of $x$".
Proof 2: Proof that $0$ is a constant, not just a symbol placeholder
From proof 1, we have that
$y + 0 = y$
$y + (x-x) = y$
From axiom 4,
$x-x = y-y$
We know that $x-x = 0$, and therefore $y-y=0$ as well, so for the equation
$x+0=x$
$0$ is a constant and therefore the same for any $x$.
Proof 3: Axiom 4 can be reduced to: There is a number $-x$ such that $x + (-x) = 0$
So basically we are trying to prove that for any numbers $x$ and $y$, there exists $z$ such that $x + z = y$
Now we work backwards a little bit:
Take any two real numbers $x, z$, and let their sum be denoted by $y$
$x + y = z$
Add $(-x)$ to both sides:
$(x + z) + (-x) = y +(-x)$
Commutative Propriety
$(z+(-x)) + (-x) = y +(-x)$
Associative Propriety $z+(x+(-x))=y+(-x)$
$x+(-x) = 0$
$z+0=y+(-x)$
$z = y+(-x)$
So $z$ exists as is denoted by $y+(-x)$
Comments/concerns: I just realized that this proof is circular, because it is dependent on proof 1 ($x+0=x$), which is dependent on axiom 4. So I don't know what to do with this. I am aware that in this version of axiom 4 we haven't introduced the subtraction operator, and even if we did we would need to prove that $y-x=y+(-x)$