Consider the following premises. For all y_2, y_2 is not sad. For all y_2, if y_2 climbs a mountain, then y_2 is sad. For all y_2, if y_2 works, then y_2 is making tea. For all y_2, if y_2 makes tea, then y_2 is working. There is at least one y_2 for which y_2 will make tea if and only if y_2 will climb a mountain. For all y_2, if y_2 were sad, then y_2 is climbing a mountain. Can we infer the following from them? If we can, name the inference rule being used: For all y_2, y_2 will work.
 From the fact that for all y_2, if y_2 were sad, then y_2 is climbing a mountain, and that for all y_2, if y_2 climbs a mountain, then y_2 is sad we can infer that for all y_2, y_2 is sad if and only if y_2 will climb a mountain via universal biconditional introduction. From the fact that for all y_2, y_2 is sad if and only if y_2 will climb a mountain, and that for all y_2, y_2 is not sad we can infer that for all y_2, y_2 will not climb a mountain via universal biconditional elimination. From the fact that for all y_2, if y_2 makes tea, then y_2 is working, and that for all y_2, if y_2 works, then y_2 is making tea we can infer that for all y_2, y_2 will make tea if and only if y_2 will work via universal biconditional introduction. From the fact that there is at least one y_2 for which y_2 will make tea if and only if y_2 will climb a mountain, and that for all y_2, y_2 will not climb a mountain we can infer that there is at least one y_2 for which y_2 will not make tea via existential biconditional elimination. Finally, from the fact that for all y_2, y_2 will make tea if and only if y_2 will work, and that there is at least one y_2 for which y_2 will not make tea we can infer that there is at least one y_2 for which y_2 will not work via existential biconditional elimination, which contradicts that For all y_2, y_2 will work. Therefore, the answer is no.