merely reflects the increasing slope of the melting curve $|dT_{\rm m}/dp|$ in accordance with the Clausisus-Clapeyron relation, as can be seen as follows. Since the free energies of lattice and liquid must be equal at $T_{\rm m}$, $$\Delta{\cal F}[T_{\rm m}(p),p]\equiv {\cal F}_{\rm lattice}[T_{\rm m}(p),p]-{\cal F}_{\rm liquid}[T_{\rm m}(p),p]=0\enspace,$$ we have, $${d\Delta{\cal F}\over dp}={\partial \Delta{\cal F}\over\partial p}+{\partial\Delta{\cal F}\over\partial T}{dT_{\rm m}\over dp} =0\enspace.
\label{eCC}$$ Since $${\partial \Delta{\cal F}\over \partial p}= L^2L_z \Delta Q\quad {\rm and}\quad
{\partial \Delta{\cal F}\over \partial T}= L^2L_z {\Delta E\over T_{\rm m}}\enspace,$$ substituting into Eq. (\[eCC\]) then gives the Clausius-Clapeyron relation for our system, $${dT_{\rm m}\over dp} = -{T_{\rm m}\Delta Q\over \Delta E}\enspace.
\label{eCC2}$$ Fitting our data (blue solid circles in Fig. \[f1\]) for $T_{\rm m}(p)$ to a quadratic polynomial in $p^2$ (solid blue line in Fig. \[f1\]), we used the fitted polynomial to determine the slope $dT_{\rm m}/dp$, and in Fig. \[f5\]b we plot $|dT_{\rm m}/dp|$ and the disorder averaged $[T_{\rm m}\Delta Q/\Delta E]$ vs disorder strength $p$. The disorder average is over 8 independent realizations of the quenched randomness. We find excellent agreement with Eq. (\[eCC2\]), thus verifying that our results are indeed very well equilibrated.

Experiment evidence for the [*absence*]{} of an upper critical point, in agreement 