By the bayes theorem is true the following statement?
$P(S | X ,U) \propto P(X | S, U)P(S|U)$
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Well, it depends on what you mean by "proportional to". In the trivial sense, yes, you can divide one expression by the other, and for any given distribution you'll get some number. The constant will not be independent of the probability distribution, though.
We can explicitly calculate it with a conditional variant of Bayes rule, with $X$ and $Z$ conditional on $Y$, or equivalently, expanding the expression $P(X, Z | Y)$ in two different ways:
$$P(X | Z,Y) P(Z | Y) = P(X, Z | Y) = P(Z | X, Y) P(X | Y)$$
So $P(Z | Y)$ (or its inverse) is the constant of proportionality, but this is different for different distributions.