Proof

Assume the equation $x\cdot z=y\cdot z$ holds and that $z\neq 0$¹.
By definition of rational numbers, each rational number $x,y,z\in\mathbb Q$ is an equivalence class of ordered pairs of some integers $a,b,c,d,e,f\in\mathbb Z$, with $b\neq 0,d\neq 0,f\neq 0$² and $x=\frac ab,$ $y=\frac cd,$ and $z=\frac ef.$
It follows by definition of multiplying rational numbers $$x\cdot z=y\cdot z \Leftrightarrow \frac {ae}{bf}=\frac {ce}{df}.$$
Again, by definition of rational numbers $$\Leftrightarrow (ae)(df)=(ce)(bf).$$
By associativity law for multiplying integers $$\Leftrightarrow aedf=cebf.$$
By commutativity law for multiplying integers $$\Leftrightarrow adef=cbef.$$
By cancellation law for multiplying integers, where $ef\neq 0$, since $z\neq 0$, by hypothesis, $$\Leftrightarrow ad=cb.$$
By definition of rational numbers $$\Leftrightarrow \frac ab=\frac cd\Leftrightarrow x=y.$$
The above arguments show the right cancellation property.
Because the multiplication of rational numbers is commutative, the left cancellation property follows immediately.
Altogether, we have shown that the multiplication of rational numbers is cancellative: $$x\cdot z=y\cdot z\Longleftrightarrow x=y,$$ for all $x,y,z\in\mathbb Q$ with $z\neq 0.$

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Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013