Page:O. F. Owen's Organon of Aristotle Vol. 1 (1853).djvu/114

 with every S, but P with a certain S, P must necessarily be present with a certain R, for since the affirmative is convertible, S will be present with a certain P, so that since R is present to every S, and S with a certain P, R will also be present with a certain P, wherefore also P will be present with a certain R. Again, if R is present with a certain S, but P is present with every S, P must necessarily be present with a certain R, for the mode of demonstration is the same, and these things may be demonstrated like the former, both by the impossible, and by exposition. If however one be affirmative, and the other negative, and the affirmative be universal, when the minor is affirmative there will be a syllogism; for if R is present with every S, and P not present with a certain S, P must also necessarily not be present with a certain R, since if P is present with every R, and R with every S, P will also be present with every S, but it is not present, and this may also be shown without deduction, if some S be taken with which P is not present. But when the major is affirmative there will not be a syllogism, e. g. if P is present with every S, but R is not present with a certain S; let the terms of being universally present with be "animate," "man," "animal." But it is not possible to take the terms of universal negative, if R is present with a certain S, and with a certain S is not present, since if P is present with every S, and R with a certain S, P will also be present with a certain R, but it was supposed to be present with no R, therefore we must assume the same as in the former syllogisms. As to declare something not present with a certain thing is indefinite, so that also which is not present with any individual, it is true to say, is not present with a certain individual, but not being present with any, there was no syllogism, (therefore it is evident there will be no syllogism).