In the second figure the syllogism is, all A
is B, no C is B,..no C is A. If proof of this is required, plainly
it may be shown either in the first figure as above, in the second
as here, or in the third. The first figure has been discussed, and
we will proceed to display the second, proof by which will be as
follows: all B is D, no C is D..., since it is required that B
should be a subject of which a predicate is affirmed. Next, since D is
to be proved not to belong to C, then D has a further predicate
which is denied of C. Therefore, since the succession of predicates
affirmed of an ever higher universal terminates, the succession of
predicates denied terminates too.
The third figure shows it as follows: all B is A, some B is not C.
Therefore some A is not C. This premiss, i.e. C-B, will be proved
either in the same figure or in one of the two figures discussed
above. In the first and second figures the series terminates. If we
use the third figure, we shall take as premisses, all E is B, some E
is not C, and this premiss again will be proved by a similar
prosyllogism. But since it is assumed that the series of descending
subjects also terminates, plainly the series of more universal
non-predicables will terminate also.
Pages:
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73