I am not sure if my proof that$(x^{-1})^{-1} = x$ using only the axioms of real numbers is correct

Question

I'm not sure if I'm allowed to do this since I am manipulating both sides at the same time( since I am starting with what I'm supposed to prove). I'm pretty sure a correct proof is when you start with the left side and then show that it's equal to the right side, but in this case, I had no idea how to do this.
$(x^{-1})^{-1} = x$
$(x^{-1})^{-1}*x^{-1} = x*x^{-1}$ (Using the property of Replacement)
1 = 1 ( Axiom of Multiplicative Inverse)

Is this proof valid ? and if by chance this proof is valid, is there a better way to prove this ?

Answer 1

We can just start with $x$ and proceed directly. $$x=1x=((x^{-1})^{-1}x^{-1})x=(x^{-1})^{-1}(x^{-1}x)= (x^{-1})^{-1}1=(x^{-1})^{-1}$$

Answer 2

Suppose that $x\in\mathbb{R}$ and $x\neq 0$.

Let us suppose there are two inverses of $x$, which we shall denote by $y$ and $z$.

Then one concludes they are equal. Indeed, one has: \begin{align*} y = y\cdot 1 = y(xz) = (yx)z = 1\cdot z = z \end{align*} which confirms its uniqueness.

Based on such result, we are now able to prove the desired claim. Indeed, one has: \begin{align*} xx^{-1} = 1 = x^{-1}(x^{-1})^{-1} = (x^{-1})^{-1}x^{-1} \Rightarrow x = (x^{-1})^{-1} \end{align*}

Hopefully this helps!