# Methods of Proof: “Perfect” Deductions, Conversion, Reduction

Methods of Proof: “Perfect” Deductions, Conversion, Reduction

Aristotle’s proofs can be divided into two categories, based on a distinction he makes between “perfect” or “complete” (teleios) deductions and “imperfect” or “incomplete” (atelês) deductions.

A deduction is perfect if it “needs no external term in order to show the necessary result” (24b23-24), and it is imperfect if it “needs one or several in addition that are necessary because of the terms supposed but were not assumed through premises” (24b24-25).

The precise interpretation of this distinction is debatable, but it is at any rate clear that Aristotle regards the perfect deductions as not in need of proof in some sense.

For imperfect deductions, Aristotle does give proofs, which invariably depend on the perfect deductions.

Thus, with some reservations, we might compare the perfect deductions to the axioms or primitive rules of a deductive system. [nb]

In the proofs for imperfect deductions, Aristotle says that he “reduces” (anagein) each case to one of the perfect forms and that they are thereby “completed” or “perfected”.

These completions are either probative (deiktikos: a modern translation might be “direct”) or through the impossible (dia to adunaton).

A direct deduction is a series of steps leading from the premises to the conclusion, each of which is either a conversion of a previous step or an inference from two previous steps relying on a first-figure deduction.

Conversion, in turn, is inferring from a proposition another which has the subject and predicate interchanged.

Specifically, Aristotle argues that three such conversions are sound:

– Eab → Eba
– Iab → Iba
– Aab → Iba

He undertakes to justify these in An. Pr. I.2.

From a modern standpoint, the third is sometimes regarded with suspicion.

Using it we can get

– Some monsters are chimeras

from the apparently true

– All chimeras are monsters;

but the former is often construed as implying in turn

– There is something which is a monster and a chimera, and thus that there are monsters and there are chimeras.

In fact, this simply points up something about Aristotle’s system:

Aristotle in effect supposes that all terms in syllogisms are non-empty.

(For further discussion of this point, see the entry on the square of opposition).

As an example of the procedure, we may take Aristotle’s proof of Camestres. He says:

If M belongs to every N but to no X, then neither will N belong to any X. For if M belongs to no X, then neither does X belong to any M; but M belonged to every N; therefore, X will belong to no N (for the first figure has come about). And since the privative converts, neither will N belong to any X. (An. Pr. I.5, 27a9-12)

[
<a,e,e> (Camestres)
M belongs to every N / every N is M
M belongs to no X / no X is M

every N is M
no X is M = no M is X [<e,a,e> (Celarent)]
no N is X =no X is N
]

From this text, we can extract an exact formal proof, as follows:

Step Justification Aristotle’s Text

1. MaN If M belongs to every N
2. MeX but to no X,

To prove:
NeX then neither will N belong to any X.

1. MeX (2, premise) For if M belongs to no X,
2. XeM (3, conversion of e) then neither does X belong to any M;
3. MaN (1, premise) but M belonged to every N;
4. XeN (4, 5, Celarent) therefore, X will belong to no N (for the first figure has come about).
5. NeX (6, conversion of e) And since the privative converts, neither will N belong to any X.

A completion or proof “through the impossible” shows that a certain conclusion follows from a pair of premises by assuming as a third premise the denial of that conclusion and giving a deduction, from it and one of the original premises, the denial (or the contrary) of the other premises.

This is the deduction of an “impossible”, and Aristotle’s proof ends at that point.

[
<o,a,o> (Bocardo)
Some F is not G. / not all F are G. / some G is not F.
Every F is J.
Some G is not J.

An example is his proof of Bocardo in 27a36-b1:

Step Justification Aristotle’s Text

1. MaN Next, if M belongs to every N,
2. MoX but to no X,

To prove: NoX then it is necessary for N not to belong to some X

1. NaX Contradictory of the desired conclusion For if it belongs to all,
2. MaN Repetition of premise 1 and M is predicated of every N,
3. MaX (3, 4, Barbara) then it is necessary that M belongs to every X.
4. MoX (5 is the contradictory of 2) But it was assumed not to belong to some.