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I'm just reading about Aristole's work on syllogism in Anthony Kenny's A New History of Philosophy book. I'm struggling to understand the tranformations between different forms of syllogisms.

So considering a basic syllogism.

All Greeks are humans.
All humans are mortal.
All Greeks are mortal.

The the terms would be

Minor (S)- Greeks
Middle (M) - humans
Major (P) - mortal

So there are three forms of syllogisms.

Form 1: S-M, M-P ⊢ S-P
Form 2 M-S, M-P ⊢ S-P
Form 3 S-M, P-M ⊢ S-P

In Prior Analytics Aristotle says that all 2nd and 3rd forms can be transformed into the ('perfect'?) 1st.

So a worked example of a transformation would be

No Greek is a bird, 
All Ravens are birds, 
No Greek is a raven;

which is in the 3rd form:

S - M
P - M
S - P

Converting the minor premise to No bird is a Greek it becomes (rearranging the premises):

All Ravens are birds, 
No bird is a Greek,
No Greek is a raven;

Which is of the 1st form.

The question

Aristole shows that almost all can be converted in this manner. However some can't be transformed in this way and have to be transformed by a process of reductio ad absurdum. Can someone give me a worked example of a syllogism that can't be transform as in my example and how it is transformed by reductio ad absurdum

Thank you if you managed to read down this far and I appreciate your expertise.

  • See the SEP entry regarding Aristotle's Logic for details. Basically, the first form is called "perfect" because the inference it express are more "intuitive" (e.g.if all A are B and all B are C, it is natural to conclude that all A are C). – Mauro ALLEGRANZA May 3 '14 at 21:42
  • About the proof “through the impossible” , can you prove Bocardo in a way different from Aristotle's example in the SEP's entry ? – Mauro ALLEGRANZA May 3 '14 at 21:45
  • I've added a proof of the fact that Bocardo cannot be proved directly. Hope it's helpful. – Hunan Rostomyan May 4 '14 at 18:14
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There are two main approaches to syllogistic proofs: proof-theoretic vs semantic.

  • The proof-theoretic route would be to consider some set of axioms for the assertoric (that is, ignoring the modalities) syllogistic, along with the so-called conversion rules (inference rules, in modern parlance), and assuming the axioms, to derive conclusions by transforming steps according to those conversion rules.

  • The semantic route would be to interpret the four forms of syllogistic proposition within set theory, reducing proofs of syllogistic consequence to very easy proofs in set theory.

I'll take this second, semantic route. I will not attempt to prove any meta-theoretic results about the syllogistic, because I'm not qualified; for an accessible exposition of that take a look at Jonathan Lear's Aristotle and Logical Theory. Instead, I will work through an example syllogistic consequence using the method of indirect proof (reductio ad absurdum), from a semantical point of view. Consider (27a36-b1):

Bocardo: MaN, MoX ⊢ NoX.

Those unfamiliar with the strange notation shouldn't get uncomfortable, because once I present the interpretation of 'a', 'o' (and there are also 'e' and 'i' relations), Bocardo will obtain an intuitive meaning:

Definition 1. (Semantics)

  1. AaB  =df         B ⊆ A              ('all Bs are As'),

  2. AeB  =df         B ∩ A = ∅       ('no Bs are As'),

  3. AiB   =df       ¬(B ∩ A = ∅)     ('some B is A'),

  4. AoB  =df       ¬(B ∩ A = B)     ('some B is not A').

Now let's look at what exactly Bocardo says:

Fact 2. (N ⊆ M), ¬(X ∩ M = X) ⊢ ¬(X ∩ N = X),

i.e. if Ns are Ms, and some Xs are not Ms, then we may conclude that some Xs are not Ns. I don't want to claim that Bocardo has no direct proof (by means of axioms and conversion rules), but merely to demonstrate how its indirect proof might proceed using this interpretation.

Proof. The premises are: (N ⊆ M)1, and ¬(X ∩ M = X)2. Suppose, for contradiction, that (X ∩ N = X). This means that every X is N. Premise (2) says that some x is s.t. x ∈ X ∧ ¬(x ∈ M), which with the hypothesis tells us that x ∈ X ∧ ¬(x ∈ M) ∧ x ∈ N. But premise (1) says that all Ns are M, so since x ∈ N, it follows that x ∈ M, contradicting the result that ¬(x ∈ M). Therefore, we obtain: ¬(X ∩ N = X).   ■

That's the form that all reductions to absurdity take. If you don't find the set-theoretic semantics quite intuitive enough, there is a nice correspondence between the expressions in our semantics and Venn diagrams, so just draw the circles and get an intuitive feel for what needs to be proved. Then, proceed formally to actually prove it in some system or another. I hope this helps answer some of your question.

                                                                     Addendum

I just realized that there is a straightforward, but somewhat involved way of proving the fact that:

Fact 3. Bocardo cannot be proved directly in Aristotle's syllogistic.

To establish this fact, we have to leave the domain of semantics and look at the syllogistic from the proof-theoretic point of view. Any adequate (i.e. sound and complete) axiomatization of assertoric syllogistic will do the job (e.g. Łukasiewicz 1929, Corcoran 1973). I'll go with the system I'm more familiar with, for an exposition of Corcoran's system see Vlasits (forthcoming), §2. Here are the axioms:

Definition 4. (Proof System):

  1. XaX                       ("all Xs are Xs");

  2. XiX                        ("some X is X");

  3. XaY, YaZ ⊢ XaZ   (Barbara);

  4. YaC, YiX ⊢ XiZ     (Datisi).

Using this system we can define the relation of direct derivability as follows (Vlasits):

Definition 5. (Direct Derivability) Γ ⊢ ψ iff there is a sequence (φ1,...,φk) s.t. ¬(ψ = φk ∈ Γ) & (∀i ≤ k):

  1. φi ∈ Γ,

  2. φi is of form XaX or XiX,

  3. j,k < i s.t. φi is obtained from φj and φk using Barbara or Datisi.

Vlasits goes on to give a proof-theoretic definition of reductio in a similar way, but for our purposes, the definition of direct derivability will be sufficient. Let's now prove that Bocardo isn't directly derivable within the aforementioned proof system, i.e., that (Fact 3) is true.

Proof. If Bocardo were directly derivable within our proof system, then there would be a sequence (φ1,...,φk) meeting the requirements specified in the definition of direct derivability. Now, Bocardo has the form: MaN, MoX ⊢ NoX, so we must have, for some i, j < k, φi = MaN, φj = MoX, and φk = NoX. Looking at the axioms and rules of our proof system we observe that MoX is neither an axiom nor is reachable by an inference rule, so it couldn't be a member of any construction sequence within this proof system. That is sufficient to show that NoX cannot be the last element of a construction sequence in our proof system. That means, i.a., that NoX is not directly derivable from {MaN, MoX}, so Bocardo is not directly provable within the assertoric syllogistic.                            ■

Admittedly, the proof is only semi-formal, but I hope it will be sufficient to motivate the claim that we really need reductio to prove certain claims of the syllogistic.

                                                                     References

Corcoran, J. (1973) "A Mathematical Model of Aristotle’s Syllogistic", AGPH 55:2.
Łukasiewicz, J. (1951) Aristotle's Syllogistic: From the Standpoint of Modern Formal Logic, 2nd Edition.
Smith, R. (2004) "Aristotle's Logic", SEP, Spring, Zalta, E.N. (ed.).
Vlasits, J. (forthcoming) "Divisional Semantics for Aristotle's Assertoric Syllogistic", Aug. 22, 2012.

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Comment: If Aristolean syllogisms are your first exposure to logic, take heart. There are much easier, more intuitive ways to solve the same problems, namely predicate logic and natural deduction that you will probably learn later. For obvious historical reasons, syllogisms may be presented first in an introductory course on logic.

We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.

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    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – Hunan Rostomyan May 6 '14 at 3:47
  • It wasn't meant to be an "answer" as such. I wasn't previously able to leave actual comments, but felt that I should leave some words of encouragement for the OP and readers in similar positions. – Dan Christensen May 6 '14 at 4:40
  • I understood your intention and I think it was a useful post, even if it wasn't meant as an answer. It was just brought to my attention by the review queue because someone flagged it, so I figured if I pressed the right button and others agreed this could automatically be converted to a comment on the question. But unfortunately instead of converting it to a comment the "machine" (I know you know a few things about helpful software!) left you a not very helpful comment from me. Sorry about that. – Hunan Rostomyan May 6 '14 at 4:48

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