3

I am reading about categorical syllogism in Copi's Introduction to Logic. I was wondering what differences are between the following statements:

  • No S is P.
  • All S are not P.

Am I correct that

  • Their quantities are both universal, instead of particular?
  • Their qualities are both negative, instead of affirmative?
  • they mean the same?

Does the book answer my question somewhere?

9
  • 1
    Normally(?) they're interchangeable, though "hyperintensionally" they might not be the same. Or, then, intuitionist negation theory might not support the interchange, for example. Commented Sep 25, 2023 at 11:55
  • 1
    See Aristotle's Logic and see also this post. Yes, they are both Universal. Commented Sep 25, 2023 at 12:09
  • @KristianBerry can you elaborate?
    – Tim
    Commented Sep 25, 2023 at 12:45
  • @Mauro do they mean the same? Why are the two links in your comment the same?
    – Tim
    Commented Sep 25, 2023 at 12:46
  • 1
    They are logically equivalent in the classical logic by the de Morgan law for quantifiers: ¬∃x P(x) = ∀x ¬P(x). They do not "mean the same", at least, in the sense that they express something in two different ways (this is what having different "intension" stands for), if we did not make this distinction then all theorems of mathematics would "mean the same" as axioms of set theory. Also, there are non-classical logics where they are not even equivalent.
    – Conifold
    Commented Sep 25, 2023 at 19:18

1 Answer 1

-2

Update: post discussion with Bumble

It depends.

The propositions E ("No S is P") and A* ("All S are not-P") may or may not be logically equivalent, depending on your interpretation of the existential import of E.

That is:

  • If E does not require the existence of S to be true, then E and A* are not logically equivalent.
  • Otherwise, they are logically equivalent.

Bumble discussed this matter in another post, which provides a fuller treatment, showing three options.

Option 1. A, E, I, O all have existential import. The main problem here is that on the face of it, it breaks the square of opposition....

Option 2. A and I have existential import. This makes much better sense of the square of opposition. But it is unnatural to understand O statements as lacking existential import...

Option 3. None have existential import. On this interpretation, the issue of existence is separate from that of predication.

I am squarely in Option 2, with the qualifier that I would formulate O as "Not [all S are P]", rather than the typical "Some S are not P": the latter formulation is an affirmation (" It is the case that 'some S are not P' "), rather than a denial (" It is not the case that 'all S are P' "), and seems to imply O has existential import when it does not.

So, read my view as assuming propositions A ("All S are P") and I ("Some S are P") have existential import, but E ("No S are P" / "Not [some S are P]") and O ("Not [all S are P]") do not.


Rest of Original Post

No, they are not the same thing in traditional/Aristotelian logic, although they may both be true in some cases.

However, in the Brentano-Venn thesis, which denies existential import for universals (both positive and negative) they are; but the propositions ought to be re-written as:

  • ("No S is P") as "There is no S that is P" and
  • ("All S are not-P") also as "There is no S that is P" to make it clear what interpretation is intended.

There is a lot of confusion about this, even from the likes of Bertrand Russell, who uses the traditional Aristotelian form of these propositions:

"The proposition 'No Greeks are men' is, of course, the proposition 'All Greeks are not-men.'"

 —Bertrand Russel, The Philosophy of Logical Atomism (1918–19), Lecture 5: General Propositions and Existence

Given the labels:

  • E: No S is P.
  • A*: All S are not-P.

E is a denial, while A* is an affirmation in the Aristotelian system.

Aristotle spent the entirety of Prior Analytics I 46 discussing their difference:

In establishing or refuting, it makes some difference whether we suppose the expressions 'not to be this' and 'to be not-this' are identical or different in meaning, e.g. 'not to be white' and 'to be not-white'. For they do not mean the same thing, nor is 'to be not-white' the negation of 'to be white', but 'not to be white'.

—Aristotle, Prior Analytics I 46

I show it visually as follows: enter image description here

To say 'A*' ("All S are not-P") is to affirm something about all of S: that all S lies outside of P and is not-P as a result. 

In this case, S must clearly exist for this proposition, A*, to be true, and so the diagram for A* has two circles (S and P).

Contrarily, to assert 'E' ("No S are P") is to deny something about all of P, to claim that nothing within the confines of P is S, and so the diagram for E has only one circle (P).

You'll notice that E is true when S does not exist, while A* is false (since there is no S to apply the label "not-P" to).

The typical logic diagram for E shows two non-overlapping circles, S and P, rather than the single circle, P, I used to depict E. You can see this in Hamilton's Lectures on Logic: enter image description here

I would argue two non-overlapping circles show A* (All S is not-P) rather than exclusively E (No S is P)—of course, E is still true in the case of A* being true—and that this construction further confuses people as to the difference between E and A*.

9
  • 2
    No, I wrote this and created the images entirely myself. I did not intend to be flashy, merely to visually depict the difference between propositions of type A* (All S are not-P) and E (No S are P). IMHO, these propositions are frequently confused and I consider the typical diagrammatic depiction of E (two non-overlapping circles) to be a part of this confusion—since such a diagram is more specifically a depiction of A*. Frankly, I'm a little surprised at the down-votes, but at least you have the good grace to give your reason.. Commented Jun 10 at 10:00
  • 2
    I'm not one of the downvoters, but I don't really like the answer. You are presenting one particular take on Aristotle's syllogistic, and one that has several problems. It is standard in Aristotelian logic for "No S is P" to entail "No P is S" by conversion. Unless you reject this, you cannot make "No S is P" asymmetrical in the way you present. Also, if you wish to extend the same reasoning to I and O statements, i.e. to hold that I has existential import and O does not, then you run into the problem that "some S is not P" standardly entails "some S is not-P" by obversion.
    – Bumble
    Commented Jun 10 at 18:28
  • 1
    @Bumble appreciate the constructive feedback. I had not considered conversion or obversion, so thanks for raising. On reflection, my interpretation does not seem to affect conversion for E. If E is true and S does not exist, Econv (No P is S) is true; if S exists, and so E & A* are true, then Econv remains true. For Econv to be false, there would have to be some overlap between S and P, but that is not given by E or E & A*. Have I misunderstood your argument here? Commented Jun 11 at 0:01
  • 1
    @Bumble your comments regarding obversion were an enjoyable challenge, but after working out the consequences, I conclude that obversion requires existence. Given E ("No S are P"), if we are to assert the obverse, A* ("All S are not-P"), then we must assume S exists (as per my Euler diagram of A* above), as otherwise S does not exist, and so the predicate not-P cannot apply to all S, since no S exists (and A* is false). Again, I'm most interested in reading your views on this point (that obversion requires existence). Commented Jun 11 at 0:10
  • 3
    In the standard way that Aristotelian logic is presented and taught, conversion and obversion apply without restriction. They are called forms of immediate inference. Also it seems plausible to consider "no S is P" and "no P is S" to be logically equivalent, and also "some S is not P" and "some S is not-P" to be logically equivalent. So to add in existence requirements appears rather arbitrary. I wrote an answer to a similar question here that gives a broader view.
    – Bumble
    Commented Jun 11 at 2:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .