every C, and C with every B, for thus there will be the impossible. And this is the first figure, if A is with every C, and C with every B. Likewise if it is demonstrated to be present with a certain one, for the hypothesis was that A was with no B, but A was assumed present with every C, and C with a certain B, but if the syllogism should be negative, the hypothesis was that A is with a certain B, for A was assumed to be with no C, and C with every B, so that there is the first figure. Also if in like manner the syllogism is not universal, but A is demonstrated not to be with a certain B, for the hypothesis was that A is with every B, but A was assumed present with no C, and C with a certain B, for thus there is the first figure.
Again, in the third figure, let A be shown to be with every B, therefore the hypothesis was that A is not with every B, but C has been assumed to be with every B, and A with every C, for thus there will be the impossible, but this is the first figure. Likewise also, if the demonstration is in a certain thing, for the hypothesis would be that A is with no B, but C has been assumed present with a certain B, and A with every C, but if the syllogism is negative, the hypothesis is that A is with a certain B, but C has been assumed present with no A, but with every B, and this is the middle figure. In like manner also, if the demonstration is not universal, since the hypothesis will be that A is with every B, and C has been assumed present with no A, but with a certain B, and this is the middle figure.
It is evident then that we may demonstrate each of the problems through the same terms, both ostensively and through the impossible, and in
