Consider the following premises. For all x_3, x_3 will go running or x_3 is not a carpenter. For all x_3, if x_3 were a carpenter, then x_3 is running. For all x_3, if x_3 goes running, then x_3 is a carpenter. There is at least one x_3 for which x_3 is not a carpenter. Can we infer the following from them? If we can, name the inference rule being used: For all x_3, x_3 will go running.
 From the fact that for all x_3, x_3 will go running or x_3 is not a carpenter we can infer that for all x_3, if x_3 were a carpenter, then x_3 is running via universal rewriting. From the fact that for all x_3, if x_3 were a carpenter, then x_3 is running, and that for all x_3, if x_3 goes running, then x_3 is a carpenter we can infer that for all x_3, x_3 is a carpenter if and only if x_3 will go running via universal biconditional introduction. From the fact that for all x_3, x_3 is a carpenter if and only if x_3 will go running we can infer that for all x_3, if x_3 goes running, then x_3 is a carpenter via universal biconditional elimination. From the fact that for all x_3, if x_3 were a carpenter, then x_3 is running, and that for all x_3, if x_3 goes running, then x_3 is a carpenter we can infer that for all x_3, x_3 is a carpenter if and only if x_3 will go running via universal biconditional introduction. Finally, from the fact that for all x_3, x_3 is a carpenter if and only if x_3 will go running, and that there is at least one x_3 for which x_3 is not a carpenter we can infer that there is at least one x_3 for which x_3 will not go running via existential biconditional elimination, which contradicts that For all x_3, x_3 will go running. Therefore, the answer is no.