Let $(X,\mathcal U)$ be a uniform space and let $U\in \mathcal U$. Is this statement true?
$$\forall V\in \mathcal U, \exists W\in \mathcal U, U\circ W\subseteq V\circ U$$


I think if the above statement is true then we can easily prove:
$$\forall V\in \mathcal U, \exists W\in \mathcal U, W\circ U\subseteq U\circ V$$



For any entourage $U,V$  there's an entourage $W$ such that $U\circ W\subseteq V\circ U$
The second statement follows from the first one by passing to inverses (i.e. reflecting along the diagonal).


Now consider the uniform structure on $\mathbb{R}$ consisting of all subsets of $\mathbb{R}^2$ that contain an open neighborhood of the diagonal. 


Just to avoid confusion with left and right, I want to use the notation
$U\circ V = \{(x,z)| \exists y: (x,y)\in U \wedge (y,z)\in V\}.$


Let $U:=\{(x,y)| |y-x|<1 \}\cup \mathbb{R}\times \{0\}$.
$V:=\{(x,y)| |y-x|<1 \}$. 
Then $V\circ U=\{(x,y)| |y-x|<2 \}\cup \mathbb{R}\times \{0\}$. 


Now let $W$ be any entourage and let $Z$ be some open neighborhood of $0$ such that $\{0\}\times Z\subset W$. Then $U\circ W$ will contain $\mathbb{R}\times Z$. Thus $U\circ W$ will not be contained in $V\circ U$. Since $W$ was arbitrary this should be a counterexample.


