Logic: proving the valid syllogisms


Added later: there is an Edit near the very end; I removed some silly text.

In this post I want to “finish off” syllogisms. No, this is certainly not the last word on syllogisms, and is almost certainly not my last word on them… but when I pick them up next, it will probably be to look at them using Boolean algebra.

What I propose to do in this post is

  • prove the four valid syllogisms in figure 1;
  • reduce the number of essentially distinct valid syllogisms from 15 to 8;
  • reduce that number from 8 to 6;
  • prove the two additional valid syllogisms;
  • suggest practical guidelines.

Let me begin by summarizing some of the information we will need to have at our fingertips.

I will be using figures 1 through 4 (the traditional, or medieval, approach) rather than figures 1 through 3 and inversions of figure 1 (Aristotle’s approach); the following shows figures 1-4, which I will use; and the inversion, which I will not. I discussed syllogisms here, and also here.

In addition, of course, we will need the traditional names, where the 3rd line, for example, names the valid figure 3 syllogisms; and I list only the 15 syllogisms considered valid nowadays. (You may recall that the change of viewpoint is called existential import: we may say that all unicorns are white even though there are no unicorns; Aristotle would have inferred that at least one unicorn exists.)

Barbara, Celarent, Darii, Ferio
Cesare, Camestres, Festino, Baroco
Disamis, Datisi, Bocardo, Ferison
Camenes, Dimaris, Fresison.

Next, we will be using the rules for dropping and adding quantifiers, and the rules of inference and the useful tautologies. For reference, here is my short form summary of Hummel’s restrictions on adding and dropping quantifiers, which appears to be all we will need.

Finally, we will need a couple of things called

  • the rules of conversion;
  • the rules of opposition.

I’ll describe them when we get there (reducing the number of essentially distinct valid syllogisms from 15 to 8 and then to 6).

prove the four syllogisms in figure 1

the rules of conversion

swapping terms

Let us consider Festino, figure 2.

The rules of conversion tell us that we may swap terms in either or both of the two premises, without changing the premises. Can we do that in such a way as to end up in figure 1, with Ferio?

Sure: just swap the terms in the first premise.

No M is P
some S is M
so some S is not P.

OK, can we systematize that?

Sure. (This notation comes from Halmos & Givant, but, in fact, Aristotle took the four valid syllogisms in figure 1 as axioms, and effectively proved all the others by converting them to figure 1. That is, this idea not new, although the notation is.)

First, as we have done earlier for I and E, and for all four in an earlier post, denote our four possible assertions by

If we denote the first premise as F1(M,P), the second premise as F2(S,M), and the conclusion as F3(S,P), then any possible syllogism in figure 1 is of the form


where F1, F2, and F3 are each one of A, I, E, O.

Further, any possible syllogism in figure 2 is of the form


(where F1, F2, and F3 are each one of A, I, E, O).

Now, rather than keep track of F1(M,P) for figure 1, and F1(P,M) for figure 2, define

F*(p,q) = F(q,p).

(That is, instead of having to read the order of M and P, all I have to look for is a star.)

Then any possible syllogism in figure 2 is of the form


That is, it has a star on the first premise.

Any possible syllogism in figure 3 is of the form


That is, it has a star on the second premise.

And yes, any possible syllogism in figure 4 is of the form


with stars on both premises.

The rules of conversion…

E(q,p) = E(p,q)
I(q,p) = I(p,q)

can now be written

E = E*
I = I*.

Let’s return to Festino in figure 2. We write it as…E*IO but E* = E, so we get EIO, figure 1, Ferio. That symbolic trick is just a simpler way to describe exactly what we did before.

But now grab Ferison in figure 3… write it as EI*O, but I* = I, so we get EIO, figure 1, Ferio. That is, again, we swap terms in the I premise.

And of course, Fresison in figure 4 is E*I*O, but E* = E, I* = I, so we get EIO, Ferio again.

It is traditional to refer to that as “per simplice” but I prefer to say that I am “swapping” the terms in a statement.

You might have noticed that the names of all four syllogisms began with “F”; that’s no accident. It tells us that the result of our manipulations will be Ferio.

What you might not have noticed is that the three names not in figure 1… have an “s” in them. It tells us that the traditional proof of each is to swap the terms, as appropriate.

There are a couple of ways to go from here. Let’s take both!

First way… we should expect that everything beginning with C is related to Celarent, everything beginning with D is related to Darii, and everything beginning with B is related to Barbara. And sure enough, for C and D, we see the same vowlels: all the D’s involve A and I, concluding in I; all the C’s involve E and I, concluding in E. But for the two B’s not in figure 1, we have A and O, concluding in O.

Note that the order of the first two vowels, i.e. the order of the premises, can be different.

We should also expect that Cesare — with the same vowel order as Celarent, and an “s” in it — can be converted to Celarent by swapping the terms in the E premise.

Let’s try it. We have a statement in figure 2, so star the first premise… E*AE, but E* = E, so we get EAE, figure 1, Celarent.

We should also expect that Datisi in figure 3 — having the same vowel order as Darii, and and “s” in it — can be converted to Darii by swapping terms in the I premise.

Let’s do it. We have a statement is figure 3, so star the second premise… AI*I, but I* = I, so we get AII, figure 1, Darii.

The other way to have done that, and a fine way to summarize it, is to write our valid syllogisms as

A*E*E, I*A*I, E*I*O.

By replacing E* by E and I* by I, we get


By deleting repeated entries of figure 1, we get


You might notice that A*EE on line four is now figure 2, and IA*I on line four is now figure 3; there’s nothing left in figure 4. In fact, both entries on line 4 are duplicates of entries on earlier lines, so we could reduce our table to


That is what Halmos and Givant get: 8 rather than 15 syllogisms.

We can go a little further.

moving premises

That has the effect of interchanging S and P — that is, we put a star on the conclusion! On the other hand, we’re still in figure 2, so we should have a star on the first premise. Well, no star means F1(M,P), and we have F1(P,M), so yes, we still have a star on the first premise… And a star on the conclusion.

Note that keeping the star on the first premise, after we interchange them, is equivalent to starring them when we do the interchange.

The same thing should happen in figure 3: interchanging the premises will leave us in figure 3, with a star on the second premise, and on the conclusion.

But for figure 4, interchanging the premises will move us into figure 1, so the two stars disappear. Except that what we’ve already done had the effect of converting the figure 4 syllogisms to other figures. We don’t have to worry about either figure 1 or figure 4, just figures 2 and 3.

OK, let’s try this.

Our list of 10 syllogisms is


(Yes, I have chosen to keep the two duplicates on line 4. Note that the rules for figures 2 or 3 apply to them, even though they’re on line 4.))

Let’s interchange premises, and star all the conclusions, for lines after the first.

E*AE*, O*AO*
AI*I*, AO*O*,
E*AE*, AI*I*.

Now replace E* by E and I* by I:


Finally, remove all the now-duplicated figure 1 syllogisms.


Fine, but why didn’t they tell us to interchange the premises?

They did. It’s “m” for move: Camestes, Disamis, Camenes, Dimaris.

And the last two of those were the remaining entries on line 4. Doing them all together, we could deal with Camenes and Dimaris implicitly — what fixes Camestres fixes Camenes, what fixes Disamis fixes Dimaris; but Camenes by itself tells us how to fix it itself: “m” — move the premises, and “s” — swap the terms in I.

I didn’t say that very well. If I used the list of 8 syllogisms, this trickery reduces it to 6. But if I had been looking at just Camenes in the original list of 15, its “m” and “s” tell me that this trickery would have converted it to a figure 1 syllogism. I worked with the list of 10 syllogisms in order to make it clear that this trickery takes care of Camenes and Dimaris.

the rules of opposition

The rules of opposition say that

A(p,q) = ~O(p,q)
I(p,q) = ~E(p,q)

We’ve seen them before: this is the duality between \exists\ and \forall\ . In words,

“All p is q” is equivalent to “It is false that some p is not q.”
“Some p is q” is equivalent to “It is false that no p is q”.

In symbols,

\forall x\ Px \Rightarrow Qx \equiv \neg \exists x\ Px \land \neg Qx
\exists x\ Px \land Qx \equiv \neg \forall x\ Px \Rightarrow \neg Qx

Baroco and Bocardo

We are left with Baroco and Bocardo. (I found the proof of Baroco out here. It’s worth looking at.) I’m still looking for an alternative to the following, but I’m not at all sure I’ll find one, since these appear to be the traditional proofs, even though I dress them in modern notation.

Let’s look at Baroco, figure 2. First, can we do a straight-up proof? Sure. (“P” in the rightmost column of lines 1 and 2 stands for “Premise”. I should have spelled it out.)

Can we do a proof by contradiction? This time, it’s “c” for contrary. It will help (me, if not you!) if we use notation.



we wish to show O(S,P).

Suppose to the contrary that O(S,P) is false. Then one of the rules of opposition…

A(p,q) = ~O(p,q)

tell us that

~O(S,P) = A(S,P), so we take our two A’s


which is figure 1, Barbara (yes, P is the middle term!), and we conclude


but then

A(S,M) = ~O(S,M)

which contradicts our second premise O(S,M). We conclude that O(S,P) cannot be false, hence it’s true. QED.

We could, in principle, have worked with the statements themselves, but using the abbreviated notation helps me move more quickly and confidently through the manipulations.

What about Bocardo?

I am certain that I can prove it directly, but I want to see the details of a proof by contradiction. (Even though I am more comfortable with the direct proof. Heck, that’s why I don’t need to see it!)

We have figure 3, so Bocardo is

Some M is not P
all M is S
so some S is not P



and we want to show


Suppose to the contrary that O(S,P) is false… then

~O(S,P) = A(S,P),

and we take the two A’s…


which is figure 1, Barbara, and we conclude


which is ~O(M,P), which contradicts our first premise.

So the names of these two syllogisms begin with “B” and contain a “c” because they can be proven, by contradiction, using Barbara.


Having reduced the number of essentially distinct syllogisms from 15 to 6, we have now proved all of those 6.

Thus, we’ve proven that these 15 syllogisms are valid. We have not proven that the other 256 – 15 = 241 are invalid. Feel free; I’m probably not going to.

But how should we use them?

Here’s one approach. Suppose we are given two premises which are each of the form A, E, I, or O. Consider the set of the two symbols associated with the two premises.

First, if the set of symbols is not one of these 5 sets…


we’re done: there is no valid conclusion. Otherwise, suppose the set is one of those 5.

Second step, first alternative: if the set of symbols is not {A,O}, then attempt to write the premises in figure 1. Swap arguments in E or I terms, and consider both possible orders of the premises.

There are three possible outcomes:

  1. we can write the premises in figure 1, and they are a valid syllogism;
  2. we can write the premises in figure 1, but they are not a valid syllogism;
  3. we cannot write the premises in figure 1.

In any case, we’re done.

Let me illustrate those three cases.

For case 1, let’s take an example from the 4-puzzles post. Suppose we are given

Please note that I am strongly suggesting that we get to order the premises as we choose. I don’t need to consider both possible orders in general — I’m just doing whatever it takes to get to figure 1, if possible.

For case 2 (note that the picture also starts case 3), suppose we are given

This is figure 2 (B is the middle term) and there is absolutely no way to convert this from figure 2 to figure 1; I would need to swap terms in an A statement, but I can’t do that. There is no valid conclusion because I cannot even get this to figure 1.

As for Baroco and Bocardo, i.e. second step, second alternative (the set of vowels is {A,O})….

The only valid syllogisms occur in figures 2 and 3… and since we cannot swap terms in A or O premises, there’s nothing we can do to alter the mood of any given pair in A and O.

So, if such a pair {A,O} is neither figures 2 or 3, then there is no valid conclusion. If such a pair is figure 2, then Baroco, AOO, is valid; but if such a pair is figure 3, then Bacardo, OAO, is valid.

Edit: I should have stopped right there. I have removed some material which followed; I wasn’t thinking clearly when I attempted to elaborate on the previous paragraph. end edit.

Are those results obvious? (Hmm. I’m not sure I’d always get them right under the lights of national TV….)

If we have figure 2… or figure 3, which I left it in the wrong order

Well, I have to be very careful with these.

Personally, I think it best to know that Baroco is figure 2, and that Bocardo is figure 3. Let’s see: the Baroque age predates Bacardi rum; or simply that Ba preceds Bo in the dictionary. (I’ve told you how bad my memory is, and how much help it needs.)

You could now try out these guidelines on the 4 logic puzzles presented here, and discussed thoroughly here.

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