I'll prove: if cost satisfies QI, then opt[i]≤opt[i+1]. This means opt is monotonic (non-decreasing): as the state index grows, the optimal split never moves backward.
Assume for contradiction that opt[i+1]=j<k=opt[i]. Since k is optimal for i: dp[k]+cost(k,i)≤dp[j]+cost(j,i). Since j is optimal for i+1: dp[j]+cost(j,i+1)≤dp[k]+cost(k,i+1). Add these inequalities.
After canceling dp[j] and dp[k]: cost(k,i)+cost(j,i+1)≤cost(j,i)+cost(k,i+1). But QI with a=j<b=k<c=i<d=i+1 says the opposite! Contradiction. So opt[i]≤opt[i+1].