Argue that [ $xy=xz \Rightarrow y=z$] on the integers without using multiplicative inverses

Question

I know this intuitively obvious, and I might be overthinking this, but is this really something you can say without arguing further? If you take for instance eigenvector eigenvalue equation [$Av=A\lambda \Rightarrow v=\lambda$] is definitely false.

I've thought about using the fact that integers have unique factorization to say that $x$ factors uniquely as does $y$ and $z$, and because $xy = xz$ the factorization of $y$ and $z$ must equal one another $\Rightarrow$ $y=z$. However this argument also relies on [ $xy=xz \Rightarrow y=z$] in the last implication.

Then I thought of showing it inductively by using $1 \cdot b= 1 \cdot a \Leftrightarrow a = b$. Asssuming [$(n-1) \cdot b= (n-1) \cdot a \Leftrightarrow a = b$]. Then observing that $n\cdot a = n\cdot b \Leftrightarrow (n-1) \cdot b+b = (n-1) \cdot a+a \Leftrightarrow ((n-1) \cdot b - (n-1) \cdot a) +b=a$ $ \Leftrightarrow a=b$ and this seems to hold.

My question is if there are any other ways to arrive at the same result?

Answer 1

$xy=xz\Leftrightarrow xy-xz=0\Leftrightarrow x(y-z)=0\Leftrightarrow x=0\lor y-z=0\Leftrightarrow x=0\lor y=z$

Edit: Thus $x\neq0\Rightarrow(xy=xz\Leftrightarrow y=z)$