**Proof:**

First show that as much as possible of the highest value/pound item must be included in the optimal solution.

Let *w*_{h}, *v*_{h} be the weight available and value of the item with the highest
value/pound ratio (item h).

Let L(i) be the weight of item i contained in the thief's loot L.

Total Value V =

If some of item h is left, and L(j) 0 for some j h, then replacing j with h will yield a higher value.

L(j)

true by definition of h