Thus: let C be the genus of A. Then, if C is not
the genus of B-for A may well have a genus which is not the genus of
B-there will be a syllogism proving A's disconnexion from B thus:
all A is C,
no B is C,
therefore no B is A.
Or if it is B which has a genus D, we have
all B is D,
no D is A,
therefore no B is A, by syllogism;
and the proof will be similar if both A and B have a genus. That the
genus of A need not be the genus of B and vice versa, is shown by
the existence of mutually exclusive coordinate series of
predication. If no term in the series ACD...is predicable of any
term in the series BEF...,and if G-a term in the former series-is
the genus of A, clearly G will not be the genus of B; since, if it
were, the series would not be mutually exclusive. So also if B has a
genus, it will not be the genus of A. If, on the other hand, neither A
nor B has a genus and A does not inhere in B, this disconnexion must
be atomic. If there be a middle term, one or other of them is bound to
have a genus, for the syllogism will be either in the first or the
second figure. If it is in the first, B will have a genus-for the
premiss containing it must be affirmative: if in the second, either
A or B indifferently, since syllogism is possible if either is
contained in a negative premiss, but not if both premisses are
negative.
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