Why does increasing the pressure cause ice to melt more?
At equilibrium, the change in Gibbs free energy is zero.  So, to maintain equilibrium, if the pressure increases, the temperature must change in tandem in such a way that the change in free energy of the solid is the same as that of the liquid.  We have, for either phase, $$\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}P$$So, $$\mathrm{d}(\Delta G)=-\Delta S \mathrm{d}T+\Delta V\mathrm{d}P=0$$where the deltas refer to the changes between liquid and solid.  We also have that, at equilibrium, $$\Delta S=\frac{\Delta H}{T}$$So, combining these equations gives $$\mathrm{d}T=\frac{T(V_\mathrm{l}-V_\mathrm{s})}{(H_\mathrm{l}-H_\mathrm{s})}\mathrm{d}P$$Since $V_\mathrm{l}&lt;V_\mathrm{s}$, this equation indicates that the equilibrium temperature for the solid-liquid transition decreases as the pressure increases.  

This shows that the melting temperature decreases. So if the pressure is increased, and the actual temperature is not changed, the temperature of the ice will be higher than the melting temperature, and the ice will melt.