Consider the following premises. For all x_9, if x_9 were a bartender, then x_9 is climbing a mountain. There is at least one x_9 for which x_9 will climb a mountain or x_9 will go running. For all x_9, if x_9 were excited, then x_9 is a bartender. For all x_9, if x_9 climbs a mountain, then x_9 is a bartender. For all x_9, if x_9 goes running, then x_9 is excited. Can we infer the following from them? If we can, name the inference rule being used: For all x_9, x_9 is not a bartender.
 From the fact that for all x_9, if x_9 climbs a mountain, then x_9 is a bartender, and that for all x_9, if x_9 were a bartender, then x_9 is climbing a mountain we can infer that for all x_9, x_9 will climb a mountain if and only if x_9 is a bartender via universal biconditional introduction. From the fact that for all x_9, x_9 will climb a mountain if and only if x_9 is a bartender we can infer that for all x_9, if x_9 climbs a mountain, then x_9 is a bartender via universal biconditional elimination. From the fact that for all x_9, if x_9 goes running, then x_9 is excited, and that for all x_9, if x_9 were excited, then x_9 is a bartender we can infer that for all x_9, if x_9 goes running, then x_9 is a bartender via universal transitivity. Finally, from the fact that for all x_9, if x_9 climbs a mountain, then x_9 is a bartender, that for all x_9, if x_9 goes running, then x_9 is a bartender, and that there is at least one x_9 for which x_9 will climb a mountain or x_9 will go running we can infer that there is at least one x_9 for which x_9 is a bartender via existential disjunction elimination, which contradicts that For all x_9, x_9 is not a bartender. Therefore, the answer is no.