Ah, you're right. If one goes to x_{i+1}, then there will be 2^i Maybes in the type. The number of Maybe-occurrences in the type doubles in each step and sharin

Ah, you're right. If one goes to x_{i+1}, then there will be 2^i Maybes in the type. The number of Maybe-occurrences in the type doubles in each step and sharing won't help there.