Question 174925
This may not be the most eloquent proof, and I'm sure there are many others, but here is one version that works:

Since G -> ~R we have that R -> ~G.
Since P -> A and A -> R and R -> ~G, we have P -> ~G.  Since G V H and P is true, then H must be true.  (Not G is true, so to make G V H true, H must be true).  Hope this helps!