[proofplan] The proof is a direct algebraic elimination in the scalar transfer-function ring. We write the three interconnection equations, solve the output equation using the inverse of $1+L$, identify the sensitivity identities $S=(1+L)^{-1}$ and $T=LS$, and then substitute the output formula into the error and controller equations. [/proofplan] [step:Write the interconnection equations] The negative-feedback convention with additive output disturbance and additive sensor noise gives the algebraic relations \begin{align*} y=Pu+d \end{align*} and \begin{align*} e=r-y-n. \end{align*} The controller equation is \begin{align*} u=Ke. \end{align*} All products are scalar transfer-function products, so multiplication commutes. The loop transfer is $L=PK$. [/step] [step:Solve for the output map] Substitute $u=K(r-y-n)$ into the plant equation: \begin{align*} y=PK(r-y-n)+d. \end{align*} Using $L=PK$, this becomes \begin{align*} y=Lr-Ly-Ln+d. \end{align*} Move the $Ly$ term to the left: \begin{align*} (1+L)y=Lr+d-Ln. \end{align*} Since $1+L$ is invertible and $S=(1+L)^{-1}$, multiplication by $S$ gives \begin{align*} y=SLr+Sd-SLn. \end{align*} In the scalar case $SL=LS=T$, so \begin{align*} y=Tr+Sd-Tn. \end{align*} This is the asserted output relation. [/step] [step:Compute the error and control maps] Use $e=r-y-n$ and the output formula just proved: \begin{align*} e=r-(Tr+Sd-Tn)-n. \end{align*} Collect terms: \begin{align*} e=(1-T)r-Sd+(T-1)n. \end{align*} Because $T=L(1+L)^{-1}$ and $S=(1+L)^{-1}$, we have \begin{align*} 1-T=(1+L)(1+L)^{-1}-L(1+L)^{-1}=S. \end{align*} Thus $T-1=-S$, and therefore \begin{align*} e=Sr-Sd-Sn. \end{align*} Finally, the controller output is $u=Ke$, so multiplying the error relation by $K$ gives \begin{align*} u=KSr-KSd-KSn. \end{align*} This proves all three closed-loop signal maps. [guided] Start from the three defining equations of the interconnection: the disturbed plant output is $y=Pu+d$, the measured feedback error is $e=r-y-n$, and the controller output is $u=Ke$. Substituting the error equation into the controller equation and then into the plant equation gives \begin{align*} y=PK(r-y-n)+d. \end{align*} With $L=PK$, this is $(1+L)y=Lr+d-Ln$. The well-posedness and invertibility assumption lets us multiply by $S=(1+L)^{-1}$, so $y=SLr+Sd-SLn$. Since the interconnection is scalar, $SL=LS=T$, and hence $y=Tr+Sd-Tn$. For the error, substitute this expression into $e=r-y-n$. The result is $e=(1-T)r-Sd+(T-1)n$. The identity $1-T=S$ follows from $T=L(1+L)^{-1}$, because $(1+L)(1+L)^{-1}-L(1+L)^{-1}=S$. Therefore $e=Sr-Sd-Sn$. Multiplying by $K$ gives $u=KSr-KSd-KSn$. [/guided] [/step]