When I have 4 random variables, $A,B,C,D$ and know that $A+B \stackrel{d}{=} C + D$ and $A \stackrel{d}{=} C$, does this imply $B \stackrel{d}{=} D$?

Going through the definition $X \stackrel{d}{=} Y$ when $P(X \leq z) = P(Y \leq z)$ for all $z \in \mathbb R$ does not seem to lead me pretty far here, but I also cannot think of a counterexample currently.

Can anyone help me with this? Thanks!



Let it be that $A$ has non-degenerate symmetric distribution and that $C=-A$.

Then $A\stackrel{d}{=}C$ and $A+0=C+2A$ (so even stronger than $\stackrel{d}{=}$) but not $0\stackrel{d}{=}2A$


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