Consider the following premises. For all x_21, if x_21 sleeps, then x_21 is an office clerk. For all x_21, x_21 will sleep if and only if x_21 is poor. For all x_21, if x_21 climbs a mountain, then x_21 is poor. For all x_21, if x_21 cooks, then x_21 is climbing a mountain. There is at least one x_21 for which x_21 will cook. Can we infer the following from them? If we can, name the inference rule being used: For all x_21, x_21 is not an office clerk.
 From the fact that for all x_21, x_21 will sleep if and only if x_21 is poor we can infer that for all x_21, if x_21 were poor, then x_21 is sleeping via universal biconditional elimination. From the fact that for all x_21, if x_21 climbs a mountain, then x_21 is poor, and that for all x_21, if x_21 were poor, then x_21 is sleeping we can infer that for all x_21, if x_21 climbs a mountain, then x_21 is sleeping via universal transitivity. From the fact that for all x_21, if x_21 cooks, then x_21 is climbing a mountain, and that there is at least one x_21 for which x_21 will cook we can infer that there is at least one x_21 for which x_21 will climb a mountain via existential modus ponens. From the fact that for all x_21, if x_21 climbs a mountain, then x_21 is sleeping, and that for all x_21, if x_21 sleeps, then x_21 is an office clerk we can infer that for all x_21, if x_21 climbs a mountain, then x_21 is an office clerk via universal transitivity. Finally, from the fact that for all x_21, if x_21 climbs a mountain, then x_21 is an office clerk, and that there is at least one x_21 for which x_21 will climb a mountain we can infer that there is at least one x_21 for which x_21 is an office clerk via existential modus ponens, which contradicts that For all x_21, x_21 is not an office clerk. Therefore, the answer is no.