Proposition: Readout and operational discard

including every untouched reference.

If $W|x{⟩}_S=|x{⟩}_S|x{⟩}_K$, then

Proof. Expand ${\rho }_{ST}={\sum }_{x,y}|x⟩⟨y{|}_S\otimes {\rho }_{xy}$. The first partial trace gives ${\sum }_x{\rho }_{xx}$. The second gives ${\sum }_x|x⟩⟨x{|}_K\otimes {\rho }_{xx}$. The calculation is unchanged by tensoring with a reference identity.