Prove that for any invertable $n\times n$ matrix A, and any $b\in\mathbb{R}^n$, there exists a unique solution to $Ax=b$ I think I've got the two ideas needed to solve this, but it feels like they're not tied together properly. I'm not sure if I'm allowed to do something like this:

Let $A$ be an invertable $n\times n$ matrix, and $b$ be an n-dimensional vector.
\begin{align}
Ax=b&\Longrightarrow A^{-1}Ax=A^{-1}b\\
&\Longrightarrow x=A^{-1}b
\end{align}
Therefore, there exists at least one solution to the equation $Ax=b$. Additionally, for the equation $Ay=b$:
\begin{align}
Ay=b&\Longrightarrow A^{-1}Ay=A^{-1}b\\
&\Longrightarrow y=A^{-1}b\\
&\Longrightarrow y=x
\end{align}
Therefore, for any two unique combinations of $A$ and $b$, there is a unique $x$ for $Ax=b$.

The problem I feel exists with this is that I'm doing two separate proofs and referencing one in the other, when I feel like I can only do that if they're combined into one single proof. Am I mistaken?
 A: A handy way to deal with uniqueness proofs is to assume by contradiction that there exist distinct solutions.
Assume that $x_1$ and $x_2$ are distinct solutions to $Ax=b$.
Then, $Ax_1 = b$ and $Ax_2 = b$. Since $A$ is invertible, we have $x_1 = A^{-1}b$ and $x_2 = A^{-1}b$. Thus, because $A^{-1}b = A^{-1}b$, we have by transitivity $x_1 = x_2$, but we assumed they are distinct.
Therefore, the solution must be unique.

This is essentially what you're trying to do, but it is not two different proofs.
Instead, we leverage the power of transitivity and reflexivity of equality to show that distinct solutions cannot exist.
A: If $A$ is invertible and $b$ is given, then $Ax=b$ iff $x = A^{-1}b$.
A: If $A$ is invertible, left multiplication by $A$ is an isomorphism on $\mathbb{R}^n$.  An isomorphism is a bijective linear map.  For the linear system
$Ax = b$, surjectivity tells us that a solution exists, and by injectivity the solution is unique.
A: Actually, your first calculation shows uniqueness, as starting from $Ax=b$ you infer that $x=A^{-1}b$. But by simply plugging in the value $A^{-1}b$ for $x$ you also get existence as for this choice of $x$ you get $Ax=AA^{-1}b=b$.
Note that the first step used the existence of a left inverse (i.e. you made use of the fact that $A^{-1}A$ is th eidentity), whereas the existence made use of the right inverse property (i.e. that $AA^{-1}$ is the identity).
