How ot proof that of $1-1=0$ and $(-1)x = -x$ holds? Is there any proof that $1-1=0$ and that $(-1)x=x$?
These proofs should be created using only multiplication, addition and several basic rules like $a+x=0$, $a+0=a$, $1a=a$ etc.
 A: Depends on what you define to be your axioms, but yeah, there are proofs.
The usual way to go about things is to first define addition as a commutative operation with a unit, and to define that unit to be called $0$. Then you define multiplication, and you define the unit of multiplication to be equal to $1$.
Subtraction, $a-b$, is usually defined as $a-b = a+(-b)$, and for each $a$, $-a$ is defined as the unique solution to the equation $a+x=0$.
Using this, you can already see that 


*

*$1-1 = 1+(-1)$ (by definition of $-$), and 

*$1+(-1)=0$ (by definition of $-1$).



You also define as your axioms that addition and multiplication are distributive, meaning that $(a+b)c=ac+bc$. Using this, you can show that  $0\cdot x + 0\cdot x = (0+0)\cdot x$ and then, using the definition of $0$, get the equation $0\cdot x+0\cdot x = 0\cdot x$ which means (after we add $-(0\cdot x)$ to both sides of the equation) that $0\cdot x = 0$ for all $x$.
Then we can go for $(-1)\cdot x$ like so:


*

*$(-1)\cdot x + x = (-1)\cdot x + 1\cdot x$ (by the definition of $1$)

*$(-1)\cdot x + 1\cdot x = (-1+1)\cdot x$ (by distributivity)

*$(-1+1)\cdot x = 0\cdot x$ (by the definition of $-1$).

*$0\cdot x=0$ (we already proved that.


So, altogether, we have the equation $$(-1)\cdot x + x = 0$$ which means, by definition, that $(-1)\cdot x = -x$.
