Let $X$ be a normal scheme. Let $s$ be a singular point and (for sake of simplicity) $X \backslash s$ regular. One says that $s$ is a rational singularity (generally: X has rational singularities if there are more then one such $s$) if $X$ is normal, of finite type over a field of characteristic zero, and there exists a proper birational map
$f: Y \to X$
from a regular scheme $Y$ such that the higher direct images of $f_{*}$ applied to $O_Y$ are trivial. That is,
$$ R^{i}f_{*}{\mathcal {O}}_{Y}=0$$
for $i>0$.
Reference: https://en.wikipedia.org/wiki/Rational_singularity or Artin paper mentioned on the linked page.
$f$ is also called resolution of singularities (of $X$) and has the property that the restriction $$f \vert _{Y \backslash f^{-1}(s)}: Y \backslash f^{-1}(s) \to X \backslash s$$ is an isomorphism.
This definition has one problem:
Does it say that there exist one such (special) resolution $f$ with $ R^{i}f_{*}{\mathcal {O}}_{Y}=0$ (that would be a stupid definition) or does the property
(*)$$ R^{i}f_{*}{\mathcal {O}}_{Y}=0$$
hold for every resolution of $X$. The statement says nothing about the uniqueness of such resolutions of singularities.
So my question is how to verify that the condition (*) is independent of choosen resolution $f: Y \to X$ of the singularity.
In other words: If one resolution has property (*), then all.