In the textbook 'Statistical Inference - an integrated approach' by Migon et. al. there is the following theorem:
$\underline{Theorem ~2.3}$
If $\mathbf{T} = \mathbf{T}(\mathbf{X})$ is a sufficient statistic for a parameter $\theta$, then $$ p(\theta ~ |~\mathbf{X}= x) = p(\theta ~ |~\mathbf{T}= t)),$$ for all priors $p(\theta)$.
The start of the proof of this theorem has me confused. It starts as follows; I will drop the bold characters and just use the lower case. The proof has the first lines,
$p(x ~ |~\theta) = p(x,t ~ | ~ \theta)$, if $t = \mathbf{T}(x)$ and $0$ if $t\neq \mathbf{T}(x)$. So,
$p(x ~|~ \theta ) = p(x~|~t,\theta)p(t ~ |~\theta)$
It is the logic of moving from this first line to the second which has me confused, firstly because it appears that the first line is saying that the value taken by $p(x|\theta)$ is dependent on whether $t = \mathbf{T}(x)$, but t doesn't appear in $p(x|\theta)$. Secondly is the restriction that $t = \mathbf{T}(x)$ which is needed for the first equality to hold carried into the expression on the second line?
It seems that it is just an application of the multiplication rule, i.e., for events $A$, $B$ and $C$ you have $$ P(A \cap B|C)= P(A|B, C)P(B|C). $$ Where $T(X)=t$ is indeed carried forward into the second equation because it means that your condition is some realization of the $T(X)$ random variable.