Proof that Cov(W+Y, Y-V) = 0 given that W, Y, and V are uncorrelated but not independent

begin{align} Cov(W+Y,Y-V)&=Cov(W,Y)-Cov(W,V)+Cov(Y,Y)-Cov(Y,V)\ &=0-0+sigma^2-0\ &=sigma^2 end{align} given that $$Cov(aW+bY,cY+dV)=ac Cov(W,Y)+ad Cov(W,V)+bc Cov(Y,Y)+bd Cov(Y,V)$$ $a=1,b=1,c=1,d=-1$ in this case, and $$Cov(Y,Y)=Var(Y)$$.

$newcommand{Cov}{operatorname{Cov}}$ The claim is false, if we make the assumption that $sigma^2not=0.$ "Uncorrelated" occurs if and only if the covariances are zero. We know that begin{align*} Cov(W+Y,Y-V) &=Cov(W,Y)-Cov(W,V)+V(Y)-Cov(Y,V)\ &=0-0+sigma^2-0\ &=sigma^2\ &not=0. end{align*} The independence of the variables is irrelevant, as are the expected values.