 Solution

The composite $g\circ f$ is proper, hence separated. If two lifts $\mathrm{Spec}R\to X$ solve a valuation-ring lifting problem for $f$, they also solve the corresponding problem for $g\circ f$. The valuative criterion for separatedness for $g\circ f$ makes them equal, so $f$ is separated.

It remains to prove existence. Start with a squareAfter composing the lower map with $g$, properness of $g\circ f$ gives a lift $a:\mathrm{Spec}R\to X$ over $Z$ whose generic restriction is $u$. The two maps $f\circ a$ and $v$ from $\mathrm{Spec}R$ to $Y$ agree on $\mathrm{Spec}K$ and have the same composite with $g$. Since $g$ is separated, its valuative uniqueness criterion gives $f\circ a=v$. Hence $a$ is the required lift for $f$.

The morphism $f$ is of finite type by hypothesis, and it is separated and satisfies valuative existence. The valuative criterion for properness therefore proves that $f$ is proper.

Solved by gpt-5.6-sol high.