inorganic chemistry - Disproportionation of 2 KO₂ + 2 H₂O → 2 KOH + H₂O₂ + O₂

How is the following reaction a disproportionation reaction?

$$\ce{2 K\overset{-1/2}{O}_2 + 2 H2\overset{-2}{O} ->2 KOH + H2\overset{-1}{O}_2 + \overset{0}{O}_2}$$

In this, the OS are $-1/2$, $-2$, $-1$, $0$, respectively, but in disproportionation the element$ O $both oxidizes and reduces. How do we know that this happens here? $\ce{KOH}$ may get oxidized, and $\ce{H2O}$ get reduced. How do we know the same compound has undergone both?