Consider the following premises. For all x_7, if x_7 plays squash, then x_7 is poor. For all x_7, if x_7 climbs a mountain, then x_7 is playing squash. For all x_7, if x_7 climbs a mountain, then x_7 is playing squash. For all x_7, if x_7 were poor, then x_7 is climbing a mountain. Can we infer the following from them? If we can, name the inference rule being used: There is at least one x_7 for which the fact that x_7 plays squash does not imply that x_7 is climbing a mountain.
 From the fact that for all x_7, if x_7 plays squash, then x_7 is poor, and that for all x_7, if x_7 were poor, then x_7 is climbing a mountain we can infer that for all x_7, if x_7 plays squash, then x_7 is climbing a mountain via universal transitivity. From the fact that for all x_7, if x_7 plays squash, then x_7 is climbing a mountain, and that for all x_7, if x_7 climbs a mountain, then x_7 is playing squash we can infer that for all x_7, x_7 will play squash if and only if x_7 will climb a mountain via universal biconditional introduction. From the fact that for all x_7, x_7 will play squash if and only if x_7 will climb a mountain we can infer that for all x_7, if x_7 plays squash, then x_7 is climbing a mountain via universal biconditional elimination. From the fact that for all x_7, if x_7 climbs a mountain, then x_7 is playing squash, and that for all x_7, if x_7 plays squash, then x_7 is climbing a mountain we can infer that for all x_7, x_7 will climb a mountain if and only if x_7 will play squash via universal biconditional introduction. Finally, from the fact that for all x_7, x_7 will climb a mountain if and only if x_7 will play squash we can infer that for all x_7, if x_7 plays squash, then x_7 is climbing a mountain via universal biconditional elimination, which contradicts that There is at least one x_7 for which the fact that x_7 plays squash does not imply that x_7 is climbing a mountain. Therefore, the answer is no.