# Chapter 22 in Organon, negative apodeictic term and problematic premise

Paragraph 4 in Chapter 22 in Prior Analytics of Aristotle:

For let us assume that A necessarily does not apply to C, and that B may apply to all C. Then by conversion of the affirmative premiss BC will give the first figure, and the negative premiss is apodeictic. But we saw that when the premisses are in this relation it follows not merely that A may not apply, but that A does not apply to some C; and so it must also follow that A does not apply to some B. When, however, the negative statement refers to the minor term, if it is problematic there will be a syllogism after substitution of the premiss, as before; but if the statement is apodeictic there will be no syllogism; for A both must apply to all B and must apply to none.

The way I understand the example is:

"A does not apply to C" and "B may apply to all C" => "A does not apply to C" and "C may apply to some B" => "A does not apply to some B".

What is the minor term in the example?

To me it looks like the negative statement is exactly apodeictic, so I don't understand how come we got a syllogism (unlike what the last sentence says).

Thanks!

• Do you intentionally omitted modalities? A's Modal Logic is quite complicated... Commented Jan 23, 2023 at 10:00
• In general, the middle term is the term occurring in both premises (and not occurring into the conclusion). In a categorical proposition, the minor is the subject while the major is the predicate: this, in "A does not apply to C", that is "no C is A", we have that C is the subject (the minor) and A is the predicate. Commented Jan 23, 2023 at 10:02
• @MauroALLEGRANZA Not intentionally omitting it...I thought that problematic and modal (e.g. possible) is the same. (I'm not really a logician, as you can tell, but I would like to understand the basics.)
– S11n
Commented Jan 23, 2023 at 15:13

See the detailed treatment of modal syllogistic into Adriane Rini, Aristotle's modal proofs: Prior Analytics A8-22 in predicate logic (Springer, 2011), page 208-on.

We have that L is necessity, M is possibility and Q is contingency.

Following modern modal logic, (possibility) is defined as ~L~ϕ, but Aristotle uses a second sense of "possible", i.e. Qϕ = ~Lϕ & ~L~ϕ (def). That is, what is contingent is neither necessary nor impossible.

Pr An, A13, 32a19: I use the expressions ‘to be possible (endechestai)’ and ‘what is possible (to endechomenon)’ in application to something if it is not necessary but nothing impossible will result if it is put as being the case.

The key points of Rini's analisys are the usual de rereadings:

LA for "It is necessary for A to belong to every B": ∀x(Bx → LAx),

and similarly for LI, LE and LO, suitable conversion rules for modalities plus so-called "CC (Complementary Conversion) rules": ∀x(Bx → QAx) is equivalent to ∀x(Bx → Q~Ax), and similar for I. [Note: in this system an assertoric (non-modal) proposition is denoted by X.]

According to Rini approach, the figure is Felapton LQX (40a25 32):

Pr1) ∀x(Cx → L~Ax)

Pr2) ∀x(Cx → QBx)

C) ∃x(Bx & ~Ax),

Here is Rini's translation:

Pr An, A22, 40a5-on: But if the privative premise is necessary, then the conclusion will be both that it is possible not to belong to some, and that it does not belong. For (1) let A be put as not belonging to C of necessity, and (2) let it be possible for B to belong to every C. Then, when the affirmative BC is converted, it will be the first figure with the privative premise necessary. But when the premises are like this, it turned out both that (C1) it is possible for A not to belong to some B and (C2) that it does not belong, so that also necessarily A does not belong to some B.

If so, we have two conlusions: C2) ∃x(Bx & ~Ax), and C1) ∃x(Bx & M~Ax), because: ϕ → Mϕ.

The second part, according to Rini's reconstruction, is as follows:

But when the privative is put in relation to the minor extreme, then if it is possible there will be a deduction when the premise is replaced, as in the previous cases [...]

The figure is called CC-Darapti LQM (40a33 35):

Pr1) ∀x(Cx → LAx)

Pr2) ∀x(Cx → Q~Bx)

C) ∃x(Bx & MAx),

and the proof is based on Darapti LQM (40a11 16), that is similar with (Pr2): ∀x(Cx → QBx) [the "reduction" to Darapti LQM is based on the "CC-conversion" of (Pr2); see above: ∀x(C → Q~Bx) is equivalent to ∀x(Cx → QBx)].

The examples chosen in the last part of the passage:

but if it is necessary no deduction can be formed. For A both necessarily belongs to every B, and cannot belong to any B. To illustrate the former take the terms sleep, sleeping horse, man; to illustrate the latter take the terms sleep, waking horse, man.

are defined as "curious" by Robin Smith [commentary to Pr An, A22, page 139].

According to Rini [page 214], the paragraph must be red as follows:

In A22 he gives only one set of Q counter-examples. These are described at 40a33-38. Aristotle is discussing why we cannot syllogize from a Q+L premise combination: ∀x(Cx → QAx) and ∀x(Cx → L~Bx), to a Q conclusion. He gives two sets of unusual terms: ‘Terms for belonging to all are sleep, sleeping horse, man; for belonging to none, sleep, waking horse, man.’

According to other readings, Aristotle's counterexamples can be fixed as follows with the triples "white, raven/swan, man": we have that "men are contingently white" [i.e. ∀x(Cx → QAx)] and we use "whiteness is necessary for swans" and "whiteness is impossible for ravens" for the two different cases of the purported conclusion: ∀x(Bx → LAx) and ∀x(Bx → L~Ax).

Thus, the minor premise with the negative: ∀x(Cx → L~Bx) will be in one case: "men cannot be swans" and "men cannot be ravens".

• Thanks! Do you have a typo in the second part: "...that is similar with (Pr2): ∀x(Cx → QBx), and the "CC-conversion" of (Pr2) to ∀x(Cx → QBx).", because you mention twice in a row "∀x(Cx → QBx)"?
– S11n
Commented Jan 28, 2023 at 9:16
• @S11n - sorry, but I cannot copy-paste all the book. The discussion is about premises translated with ∀x(Cx → QAx) and ∀x(Cx → L~Bx) (see text above). We read them as "Whiteness (A) is Contingent for Men (C)" and "Men (C) is not Possible (L~) for Swans (in one case) and Ravens (in the other case) (B)". Commented Feb 1, 2023 at 9:53
• If so, the two counterexamples are used to show that neither ∀x(Bx → LAx) nor ∀x(Bx → L~Ax) can be concluded. Commented Feb 1, 2023 at 9:56
• Nothing changes if we use the conclusion ∃x(Bx & MAx), as above, because "there is a Swan that is possibly White" is true while "there is a Raven that is possibly White" is false. Commented Feb 1, 2023 at 10:05
• @S11n - you are welcome :-) Commented Mar 1, 2023 at 8:59