Conditional independence proof I want to prove that
$\mathbb{P}(X|U,P) = \mathbb{P}(X|U) \implies \mathbb{P}(X|U,P,T) = \mathbb{P}(X|U,T)$
Where all the letters denote random variables. I'm not sure that this is right, but it seems to be an intuitive result.
Thanks.

Without further assumptions, the result does not hold.
Here's a counterexample.
Let $\mathbb{P}$ be the uniform probability on the Boolean cube $\{0,1\}^3$. Let $X,Y,Z$ be the first, the second, and the third coordinate function of the Boolean cube, respectively. Define $U:=\max(X,Y)$, $P:=Z$, and $T :=\min(Y,Z)$. Since $X,Y,Z$ are $\mathbb{P}$-independent of each other, we have
\begin{equation}
\mathbb{P}(X\mid U,P) = \mathbb{P}\big(X\mid \max(X,Y),Z\big) = \mathbb{P}\big(X\mid \max(X,Y)\big)=\mathbb{P}(X\mid U).
\end{equation}
On the other hand,

*

*$\{U=1\}\cap\{P=1\}\cap\{T=0\} = \big\{(1,0,0)\big\}$,

*$\{U=1\}\cap\{T=0\} = \big\{(1,0,0),(0,1,0),(1,1,0),(1,0,1)\big\}$,

*$\{X=1\}\cap\{U=1\}\cap\{T=0\} = \big\{(1,0,0),(1,1,0),(1,0,1)\big\}$,

from which it follows that
\begin{equation}
\mathbb{P}(X=1 \mid U=1,P=1,T=0) = 1\neq \frac{3}{4} = \frac{\frac{3}{8}}{\frac{4}{8}} = \mathbb{P}(X=1 \mid U=1,T=0),
\end{equation}
which implies $\mathbb{P}(X\mid U,P,T)\neq\mathbb{P}(X\mid U,T)$.
