Statements: a. All tigers are lions. b. All lions are horses. c. No horses are monkeys. Conclusions: I. All tigers are horses. II. No tigers are monkeys. III. Some lions are tigers. IV. Some monkeys are not tigers

From the statements, $\text{Tigers} \subseteq \text{Lions}$ and $\text{Lions} \subseteq \text{Horses}$. Therefore $\text{Tigers} \subseteq \text{Horses}$, so conclusion $\mathrm{I}$ follows. Also, no horses are monkeys, so $\text{Horses} \cap \text{Monkeys} = \varnothing$. Since all tigers are horses, tigers also cannot be monkeys. Hence conclusion $\mathrm{II}$ follows. Conclusion $\mathrm{III}$ says some lions are tigers. From $\text{Tigers} \subseteq \text{Lions}$, existence of tigers is not guaranteed, so this does not necessarily follow. Conclusion $\mathrm{IV}$ says some monkeys are not tigers. Since no horses are monkeys and all tigers are horses, no tiger can be a monkey. Thus any monkey, if it exists, is not a tiger, so this follows in standard syllogism convention. So the conclusions that follow are $\mathrm{I, II, IV}$. Re-checking the options, this matches option $\mathrm{C}$.