Proof of Theorem MIU

As described in Proof Technique U, we will assume that $A$ has two inverses. The hypothesis tells there is at least one. Suppose then that $B$ and $C$ are both inverses for $A$, so we know by Definition MI that $AB=BA=I_n$ and $AC=CA=I_n$. Then we have, \begin{align*} B &=BI_n&&\knowl{./knowls/theorem.MMIM.knowl}{\text{Theorem MMIM}}\\ &=B(AC)&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\ &=(BA)C&&\knowl{./knowls/theorem.MMA.knowl}{\text{Theorem MMA}}\\ &=I_nC&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\ &=C&&\knowl{./knowls/theorem.MMIM.knowl}{\text{Theorem MMIM}} \end{align*}

So we conclude that $B$ and $C$ are the same, and cannot be different. So any matrix that acts like an inverse, must be the inverse.