Question about posterior distributions and sufficient statistics

1. If $X_1,X_2,\ldots,X_n$ are discrete r.v.'s with joint pmf $f(x_1,\ldots,x_n|\theta)$. Let theta be a discrete random variable with prior pmf $\pi(\theta)$. Let $H(x_1,x_2,\ldots,x_n)$ be a sufficient statistic. Show that $\pi(\theta|x_1=x_1,\ldots,x_n=x_n) = \pi(\theta|H=h)$.

Attempt: Basically the posterior distribution of theta given the sample is equivalent to the posterior of theta given a sufficient statistic of the sample as a sufficient statistic contains all the relevant information about the sample.

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I'm going to write $X$ for $(X_1,\ldots,X_n)$ and $x$ for $(x_1,\ldots,x_n)$.
The likelihood function when $X$ is observed to have a certain value $x$ is $$L_1(\theta) = \Pr(X=x \mid \theta) = E(\Pr(X=x\mid H(X),\theta)) = \sum_h \Pr(H(X)=h\mid\theta)\Pr(X=x\mid H(X)=h).$$ In the very last probability, we don't need to write $\Pr(X=x\mid H(X)=h,\theta)$, since lack of this dependence on $\theta$ is the very definition of sufficiency. The index $h$ of course runs through the discrete set of all possible values of the random variable $H(X)$.
Now $\Pr(X=x\mid H(X)=h) = 0$ for any value of $h$ except $h=H(x)$. So all but one of the terms in the last sum vanish and the sum is $$\Pr(H(X)=h\mid\theta)\Pr(X=x\mid H(X)=h) =\Pr(H(X)=H(x)\mid\theta)\Pr(X=x\mid H(X)=H(x)).$$ The second factor in the last product does not depend on $\theta$. So as a function of $\theta$, it's constant. Therefore the likelihood function when $H(X)$ is observed to have a certain value $H(x)$ is $$L_2(\theta) = \Pr(H(X) = h\mid \theta) = \mathrm{constant}\cdot \Pr(X=x\mid \theta) = \mathrm{constant} \cdot L_1(\theta).$$
The "constant" factor referred to above goes away when you normalize; it appears as a factor in both the numerator and the denominator and cancels. Therefore you get the same posterior pmf if you observe $H(X)$ that you get if you observe $X$.