§1.2.12Φανερὸν οὖν ὅτι ἐν τῷ πρώτῳ σχήματι τὰ μὲν ἄλλα προβλήματα πάντα δείκνυται διὰ τοῦ ἀδυνάτου, τὸ δὲ καθόλου καταφατικὸν οὐ δείκνυται.
It is clear, therefore, that in the first figure all other problems are demonstrated through impossibility, but the universal affirmative is not demonstrated.
ἐν δὲ τῷ μέσῳ καὶ τῷ ἐσχάτῳ καὶ τοῦτο δείκνυται.
But in the middle and in the last figure this also is demonstrated.
κείσθω γὰρ τὸ Α μὴ παντὶ τῷ Β ὑπάρχειν, εἰλήφθω δὲ τῷ Γ παντὶ ὑπάρχειν τὸ Α. οὐκοῦν εἰ τῷ μὲν Β μὴ παντί, τῷ δὲ Γ παντί, οὐ παντὶ τῷ Β τὸ Γ. τοῦτο δʼ ἀδύνατον· ἔστω γὰρ φανερὸν ὅτι παντὶ τῷ Β ὑπάρχει τὸ Γ, ὥστε ψεῦδος τὸ ὑποκείμενον.
For let it be assumed that A does not belong to all B, and let A be taken to belong to all Γ. If, then, [A] does not belong to all of B, but to all of Γ, Γ does not belong to all B. But this is impossible; for let it be clear that Γ belongs to all B, so that the assumption is false.
ἀληθὲς ἄρα τὸ παντὶ ὑπάρχειν.
Therefore, it is true that [A] belongs to all [B].
ἐὰν δὲ τὸ ἐναντίον ὑποτεθῇ, συλἀογισμός μὲν ἔσται καὶ τὸ δύνατον, οὐ μὴν δείκνυται τὸ προτεθέν.
But if the contrary is assumed, there will be a syllogism and the impossibility, but what was proposed is not demonstrated.
εἰ γὰρ τὸ μηδενὶ τῷ Β, τῷ δὲ Γ παντί, οὐδενὶ τῷ Β τὸ Γ. τοῦτο δʼ ἀδύνατον, ὥστε ψεῦδος τὸ μηδενὶ ὑπάρχειν.
For if [A belongs] to no B, but to all Γ, Γ [belongs] to no B. But this is impossible, so that it is false that it belongs to none.
ἀλλʼ οὐκ εἰ τοῦτο ψεῦδος, τὸ παντὶ ἀληθές.
But it is not the case that, if this is false, belonging to all is true.
ὅτι δὲ τινὶ τῷ Β ὑπάρχει τὸ Α, ὑποκείσθω τὸ Α μηδενὶ τῷ ὑπάρχειν, τῷ δὲ Γ παντὶ ὑπαρχέτω.
And to show that A belongs to some B, let A be assumed to belong to no [B], and let it belong to all Γ.
ἀνάγκη οὖν τὸ Γ μηδενὶ τῷ Β. ὥστʼ εἰ τοῦτʼ ἀδύνατον, ἀνάγκη τὸ Α τινὶ τῷ Β ὑπάρχειν.
It is necessary, therefore, that Γ belongs to no B. So that if this is impossible, it is necessary that A belongs to some B.
ἐὰν δʼ ὑποτεθῇ τινὶ μὴ ὑπάρχειν, ταὐτʼ ἔσται· ἅπερ ἐπὶ τοῦ πρώτου σχήματος.
But if it is assumed not to belong to some, the same will happen, just as in the first figure.
πάλιν ὑποκείσθω τὸ Α τινὶτῷ Β ὑπάρχειν, τῷ δὲ Γ μηδενὶ ὑπαρχέτω.
Again, let A be assumed to belong to some B, and let it belong to no Γ.
ἀνάγκη οὖν τὸ Γ τινὶ τῷ Β μὴ ὑπάρχειν.
It is necessary, therefore, that Γ does not belong to some B.
ἀλλὰ παντὶ ὑπῆρχεν, ὥστε ψεῦδος τὸ ὑποτεθέν· οὐδενὶ ἄρα τῷ Β τὸ Α ὑπάρξει.
But it did belong to all, so that the assumption is false; therefore A will belong to no B.
ὅτι δʼ οὐ παντὶ τὸ Α τῷ Β, ὑποκείσθω παντὶ ὑπάρχειν, τῷ δὲ Γ μηδενί.
And to show that A does not belong to all B, let it be assumed to belong to all, and to no Γ.
ἀνάγκη οὖν τὸ Γ μηδενὶ τῷ Β ὑπάρχειν.
It is necessary, therefore, that Γ belongs to no B.
τοῦτο δʼ ἀδύνατον, ὥστʼ ἀληθὲς τὸ μὴ παντὶ ὑπάρχειν.
But this is impossible, so that it is true that it does not belong to all.
φανερὸν οὖν ὅτι πάντες οἱ συλλογισμοὶ γίνονται διὰ τοῦ μέσου σχήματος.
It is clear, therefore, that all the syllogisms are produced through the middle figure.
§1.2.13Ὁμοίως δὲ καὶ διὰ τοῦ ἐσχάτου.
And similarly also through the last figure.
κείσθω γὰρ τὸ Α τινὶ τῷ Β μὴ ὑπάρχειν, τὸ δὲ Γ παντί· τὸ ἄρα Α τινὶ τῷ Γ οὐχ ὑπάρχει.
For let it be assumed that A does not belong to some B, and Γ to all; therefore A does not belong to some Γ.
εἰ οὖν τοῦτʼ ἀδύνατον, ψεῦδος τὸ τινὶ μὴ ὑπάρχειν, ὥστʼ ἀληθὲς τὸ παντί.
If, then, this is impossible, it is false that it does not belong to some, so that it is true that it belongs to all.
ἐὰν δʼ ὑποτεθῇ μηδενὶ ὑπάρχειν, συλλογισμὸς μὲν ἔσται καὶ τὸ ἀδύνατον, οὐ δείκνυται δὲ τὸ προτεθέν· ἐὰν γὰρ τὸ ἐναντίον ὑποτεθῇ, ταὐτʼ ἔσται ἅπερ ἐπὶ τῶν πρότερον.
But if it is assumed to belong to none, there will be a syllogism and the impossibility, but what was proposed is not demonstrated; for if the contrary is assumed, the same will happen as in the previous cases.
ἀλλὰ πρὸς τὸ τινὶ ὑπάρχειν αὕτη ληπτέα ἡ ὑπόθεσις.
But to show that [A] belongs to some, this assumption must be taken.
εἰ γὰρ τὸ μηδενὶ τῷ Β, τὸ δὲ Γ τινὶ τῷ Β, τὸ Α οὐ παντὶ τῷ Γ. εἰ οὖν τοῦτο ψεῦδος ἀληθὲς τὸ Α τινὶ τῷ Β ὑπάρχειν.
For if [A belongs] to no B, and Γ to some B, A [does] not [belong] to all Γ. If, then, this is false, it is true that A belongs to some B.
ὅτι δʼ οὐδενὶ τῷ Β ὑπάρχει τὸ Α, ὑποκείσθω τινὶ ὑπάρχειν, εἰλήφθω δὲ καὶ τὸ Γ παντὶ τῷ Β ὑπάρχον.
And to show that A belongs to no B, let it be assumed to belong to some, and let Γ also be taken to belong to all B.
οὐκοῦν ἀνάγκη τῷ Γ τινὶ τὸ Α ὑπάρχειν.
It is necessary, then, that A belongs to some Γ.
ἀλλʼ οὐδενὶ ὑπῆρχεν, ὥστε ψεῦδος τὸ τινὶ τῷ Β ὑπάρχειν τὸ Α. ἐὰν δʼ ὑποτεθῇ παντὶ τῷ Β ὑπάρχειν τὸ Α οὐ δείκνυται τὸ προτεθέν, ἀλλὰ πρὸς τὸ μὴ παντὶ ὑπάρχειν αὕτη ληπτέα ἡ ὑπόθεσις.
But it did belong to none, so that it is false that A belongs to some B. But if it is assumed that A belongs to all B, what was proposed is not demonstrated; rather, to show that it does not belong to all, this assumption must be taken.
εἰ γὰρ τὸ Α παντὶ τῷ Β καὶ τὸ Γ παντὶ τῷ Β, τὸ ὑπάρχει τινὶ τῷ Γ. τοῦτο δὲ οὐκ ἦν, ὥστε ψεῦδος τὸ παντὶ ὑπάρχειν.
For if A belongs to all B, and Γ to all B, [A] belongs to some Γ. But this was not the case, so that it is false to belong to all.
εἰ δʼ οὕτως, ἀληθὲς τὸ μὴ παντί.
If so, it is true that it does not belong to all.
ἐὰν δʼ ὑποτεθῇ τινὶ ὑπάρχειν, ταὐτʼ ἔσται ἃ καὶ ἐπὶ τῶν προειρημένων.
But if it is assumed to belong to some, the same will happen as in the case of those previously mentioned.
Φανερὸν οὖν ὅτι ἐν ἅπασι τοῖς διὰ τοῦ ἀδυνάτου λογισμοῖς τὸ ἀντικείμενον ὑποθετέον.
It is clear, therefore, that in all reasoning through impossibility the contradictory must be assumed.
δῆλον δὲ καὶ ὅτι ἐν τῷ μέσῳ σχήματι δείκνυταί πως τὸ καταφατικὸν καὶ ἐν τῷ ἐσχάτῳ τὸ καθόλου.
And it is also evident that in the middle figure the affirmative is demonstrated in a way, and in the last the universal.