= Solution
For the <single-line greenhouse deficit>, use <hydrostatic balance>, fixed <Pressure>[surface pressure] $p_s$, constant <gravity> $g$, and the <dry adiabat>
$$
T(p)=T_s(p/p_s)^\kappa,\qquad \kappa=R_b/c_{p,b}>0.
$$
The gas constant and <Specific heat capacity>[specific heat] refer to the background gas. Let $\mu_d=\cos\tilde\theta>0$ be the supplied effective ray cosine. The vertical <optical depth> and slant <transmittance> from level $p$ to space are
$$
\tau_\nu(p)=\frac qg\int_0^p\kappa_\nu(p',T(p'))\,dp',\qquad
\mathfrak T_\nu(p)=\exp[-\tau_\nu(p)/\mu_d].
$$
This permits pressure-dependent <Spectral line broadening>[line widths]. With <local thermodynamic equilibrium>, negligible infrared <scattering>, a black thermal surface, and no incident <thermal radiation> from space, the formal <radiative transfer> solution is
$$
I_\nu(0)=B_\nu(T_s)\mathfrak T_\nu(p_s)+
\int_0^{p_s}B_\nu(T(p))[-\partial_p\mathfrak T_\nu(p)]\,dp.
$$
The weights add to one. Subtract the transparent-surface <radiance> and put $B(p)=B_{\nu_0}(T(p))$, using the narrow-line approximation only in the line deficit. Define its pressure-dependent <equivalent width>
$$
E(p,q)=\int[1-\mathfrak T_\nu(p)]\,d\nu,\qquad
\partial_pE=\int\frac{q\kappa_\nu(p)}{g\mu_d}\mathfrak T_\nu(p)\,d\nu\ge0.
$$
Approximating each hemispheric flux by $\pi$ times the effective-ray <radiance> preserves the black-surface flux. The outgoing longwave flux is therefore
$$
F_{\rm out}=\sigma_{\rm SB}T_s^4-D(T_s,q),\qquad
\boxed{D=\pi\int_0^{p_s}[B(p_s)-B(p)]\,\partial_pE(p,q)\,dp.}
$$
Integration by parts gives the particularly transparent form
$$
\boxed{D=\pi\int_0^{p_s}E(p,q)\frac{dB(p)}{dp}\,dp\ge0.}
$$
Both boundary terms vanish: $E(0,q)=0$ and $B(p_s)-B(p_s)=0$. The <dry adiabat> and monotonic <Planck function> give $dB/dp>0$. Atmospheric emission replaces part of the warm surface radiation by emission from cooler levels, reducing emission to space for a specified $T_s$. The inequality is strict when <opacity> occurs over a nonzero range with <temperature> contrast; an <Isothermal process>[isothermal] absorbing layer at the surface <temperature> would not produce such a contrast.
For a spherical <planet> the mean absorbed instellation is $F_*=(1-\alpha)L_{\circledast}/4$. The required global <energy balance> is
$$
\boxed{\sigma_{\rm SB}T_s^4=\frac{(1-\alpha)L_{\circledast}}4+D(T_s,q).}
$$
Consequently every equilibrium of this model has $T_s\ge T_{\rm bare}=[(1-\alpha)L_{\circledast}/(4\sigma_{\rm SB})]^{1/4}$, with strict warming when $D>0$. At fixed <Temperature>[temperature profile], increasing $q$ increases $E$ and hence $D$. This argument does not assume that the greenhouse gas changes the stellar-band <albedo>, <Pressure>[surface pressure], or background thermodynamics; those are held fixed. For sufficiently broad or overlapping saturated wings the full frequency-dependent <Planck function> must replace the narrow-line approximation.
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