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Validity, Venn diagrams, and *ex falso quodlibet*

(The first 2 paragraphs here are set-up; those in the know can start at the third paragraph.) I teach a class called "Critical Thinking and Reasoning" most semesters. It covers argument recognition, reconstruction, and evaluation. At the very end of the semester, we talk about how one can show a particular argument form is valid. For propositional/ sentential logic, we use truth tables: if there is no row of the truth table where all the premises are true and the conclusion false, then the argument form is valid. For categorical logic, we use Venn diagrams. If you are not familiar with how this works, it's very straightforward and simple; here is a quick introduction for the unacquainted. The basic idea is that you diagram all the premises on a single diagram, and then check whether you have 'already' diagrammed the conclusion.

One interesting thing about presenting both of these to an intro class is their difference over ex falso quodlibet: in classical propositional logic, any argument with inconsistent premises is valid, whereas in categorical logic there are invalid arguments with inconsistent premises. This is reflected in the semantics for the two logics, truth-table or Venn diagram, respectively. That is, if you diagram 'All A are B' and 'Some A are not B', you have

**not**already diagrammed 'All C are D' (or whatever -- I mean, or

*quodlibet*). Whereas, in the truth table case, if one of your premises is

*p*and another is 'not-

*p*', it is of course impossible for there to be a row where both all the premises are true and the conclusion is false -- since there is no row in which all the premises are true.

Everything so far is completely uncontroversial. Now comes the point I've been wondering about for the last couple of days. What if we set up a validity test for a propositional language that looked more like the Venn-diagram test for categorical logic? That is, instead of thinking of validity in the usual way as absence of counterexamples (in the propositional case, no row of the truth table has all true premises and a false conclusion), we demand that the diagram of all the premises be a diagram of the conclusion. In categorical logic, each of the points in the Venn diagram represents an individual object, and the circles represent sets of objects; in a propositional Venn diagram, we let each point be a case/circumstance/state of affairs/whatever, and let the circles be sets of cases etc. The resulting propositional logic would NOT be classical, since ex falso quodlibet would not hold.

Why care about setting up such a Venn-diagram validity test for propositional logic? Here's why: when relevance logicians (and anyone else who doesn't like EFQ) accept the 'no counter-example' account of validity, they are forced to say some pretty counterintuitive things, first and foremost that there is a case (or whatever) in which both

*p*and '

*not-p*' are true (and some other

*q*false). If they don't say this, then they can't say EFQ is invalid. But if this Venn-diagram validity test for a propositional language is viable, we can reject EFQ without accepting true contradictions. All that's required is tweaking the no-counterexample notion of consequence -- a tweak that is already used in critical thinking textbooks.

Finally, this idea seems so obvious that somebody must have already explored it. Any pointers to the relevant literature?

## 19 Comments:

"That is, if you diagram 'All A are B' and 'Some A are not B', you have not already diagrammed 'All C are D'"

How exactly do you represent both of this sentences in a single diagram without contradicting diagram construction rules?

Hi Rafal --

You are exactly right. (In the original draft of this post, I addressed this, but cut it out when paring it down.)

To diagram 'All A are B' and 'Some A is not B' together, you have to draw, so to speak, an impossible picture: as you say, a diagram that contradicts the diagram construction rules.

BUT: I still haven't diagrammed 'All C are D,' even if I've broken the construction rules. (Actually, as I'm saying this, I realize I don't really know anything about this diagram construction procedure beyond what's in my intro textbook -- are there special, accepted conventions in place for what happens when you have contradictory premises?)

Won't this just be Boolean logic on a propositional rather than categorial interpretation?

The text I have ready to hand gives rules for constructing diagrams that allow for inconsistency. It just means marking an area as empty and also marking it as having a member.

The text calls this an "antilogism" and suggests this method for testing a syllogism: Diagram the premises and the negation of the conclusion; the argument is valid iff the resulting diagram is an antilogism (ie, it's inconsistent). This, of course, will give you EFQ.

The book also includes the direct method of diagramming the premises and checking to see if you've diagrammed the conclusion. This, as you note, won't give you EFQ.

Thinking beyond propositional and categorial logic, the natural analog of a Venn diagram (as a semantic search procedure) is a truth tree. The usual method of constructing a tree for the premises and the negation of the conclusion (to see if every branch closes) is like the method of antilogism - gives you EFQ. It would be possible, I guess, to construct a tree for just the premises and see if you can get the conclusion in every branch. (Maybe?)

Hi Guys,

So, afaik, the usual way doesn't allow for picturing contradictory premises (come to think of it, originally the method was developed for classical syllogistics, where you have to have two premises and three terms - now, can you have two contradictory premises of this kind?).

Right, you can diagram the premises and the negation of the conclusion (once you throw in a bunch of rules that tell you how to diagram negations of assertoric sentences). Then, of course, the argument would be valid iff you end up with an antilogism. This would be a more general method than the straightforward one (also, observe, on this approach, indeed anything follows from a contradiction).

Doing direct trees... hmm ... an interesting idea. If you don't do anything that mimics an indirect proof, you may have hard time introducing variables that don't occur in the premises. So, say you wanna prove q from p and ~p. You write down p and ~p and try to derive q. How exactly are you going to introduce a new variable (recall that trees are meant to decompose your truth conditions)?

Of course, this depends on how you formulate your system. If you have disjunctive syllogism rule and vIntro you can do that (well, go to pvq first and then use ~p and pvq to obtain q). But then, it's no longer a tree, it's a natural deduction proof, no?

@PD:

"Won't this just be Boolean logic on a propositional rather than categorial interpretation?"

Yes, that's exactly what I have in mind; I just wanted to spell it out for those unfamiliar. Looking back at the original post, I see I was unclear/ misleading about what I take to be the new/interesting point here. What I meant to say is: can we use this Boolean interpretation to create a test for propositional validity that classes EFQ as invalid? (The original post is sufficiently misleading that I may go back and change it.)

More on the other stuff later.

If that's what you meant, then yeah, you can. Think of intersection as conjunction, union as disjunction, complement as negation etc. Then each subset of the domain can be in one of four "states":

- no mark in it

- marked as empty but not as non-empty

- marked as non-empty, but not marked as empty

-marked as both empty and non-empty

Think of emptiness as falsehood, non-emptiness as truth. Say you admit that all of the four above are real possibilities.

Define consequence operation thus:

G entails p

iff in any (completed) diagram where all elements of G are (at least) non-empty, p is also marked as non-empty.

Now, you have to introduce some rules for how you complete the diagram once you picture the truth of the premises. One way to go:

A^B is marked non-empty iff A is marked non-empty and B is marked non-empty.

A^B is marked empty if either A or B is marked empty.

AvB is marked non-empty iff either A or B is marked non-empty

AvB is marked empty iff both A and B are marked empty.

~A is marked non-empty iff A is marked empty

~A is marked empty iff A is marked non-empty.

If you do that, you get another decision procedure (a fancy variant of a truth-table, if you will) for FDE, first-degree entailment, which indeed, is paraconsistent in the sense that it invalidates EFQ.

Oh, I forgot to say. If you disallow the case where a set is neither empty nor non-empty, you get Priest's LP (Logic of Paradox), yet another (and a slightly stronger) paraconsistent logic.

Hi Rafal --

Thanks, that's very helpful. Do you have a reference?

Question: I'm having trouble seeing how this scheme does not make disjunctive syllogism valid. (And I thought disjunctive syllogism was invalid in FDE.)

Hey,

Check out for instance G. Priest's "An Introduction to Non-Classical Logic", esp. ch. 8 (of the second edition).

You're right, DS is invalid in FDE. How is DS invalidated? Well, consider a situation where:

P is only FALSE but it isn't TRUE

Q is both FALSE and TRUE.

(so, in terms of the diagram: you mark P as empty and you don't mark it as non-empty, and you mark Q as both empty and non-empty).

since Q is (among other things) FALSE, ~Q is (among other things) TRUE.

Also, since Q is (among other things) TRUE,

PvQ

is TRUE (it's also FALSE).

So you have

PvQ is TRUE

~Q is TRUE

but it still is not the case that P is TRUE, because it is only FALSE.

So, in short, if a disjunct is both true and false, it makes both its negation and any disjuction it occurs in true, no matter what the value of the other disjunct is.

If you want to draw a dragam, here's what to do (I use letters for sets and for variables they're associated with, I hope this is not confusing). Recall, that the distinguished value is non-emptiness. Cross all the content of P out as empty, cross all the content of Q out as empty, but put a plus sign inside of Q OUTSIDE of P. Then our rule says that ~Q is marked non-empty if Q is marked empty. So you have to put a plus sign outside of Q and P. This makes your ~Q true in this setting. Also, since the union of P and Q corresponds to PvQ and there's a plus sign in Q, there is a plus sign in the union and thus PvQ is made true. Yet, there's no plus sign within P (and if you think about the rules, no rule will force you to put a plus sign there in this situation). Having said all this, I don't think this is the most convenient decision procedure for FDE.:)

cheers,

r

Coming back to syllogistics and explosion, only now I stumbled upon an interesting passage in Prior Analytics 63b31-64a16, where Aristotle speaks about syllogisms with contradictory premises:

"...no deduction whether affirmative or negative can be made out of opposed propositions: no affirmative deduction is possible because both propositions must be affirmative, but opposites are the one affirmative, the other negative..."

Thanks again, Rafal.

That Aristotle passage is particularly helpful.

Since you wrote the long comment/proposal yesterday, I've been thinking about it -- in particular, the bit where you allow a region to be marked both empty and non-empty. That seems equivalent to what I was trying to avoid in the original post, namely accepting that some p could be both true and false.

But if one doesn't allow marking a region as empty and non-empty, then you can't really diagram categorical contradictory premises. (Which seems like a problem.)

The Aristotelian view seems to be something like: if you have to diagram contradictory premises, then throw out the diagram (at least for the purposes of inferring consequences from the premises).

Switching gears a little: this problem of diagramming impossible venn diagrams comes up when dealing with sentences like 'All A are B' and 'Some A are not B'. BUT: will it arise in the propositional case? Contradictions there are not impossible pictures, but rather ones where every world/model/case is marked empty. So it seems to me that one could, in the propositional case, disallow the possibility of some region being marked both empty and non-empty -- that is, disallow sentences being both true and false.

But I have to think about this a

lotmore before I'm willing to say anything with confidence...The point of a diagram *IS* that you can illustrate paradox or contradiction without violating the rules of self-reference. The real question is how you use the information. You read a book on feng shui. But you live in your parents basement and your broke. You cant use the info but the 5 elements and the directions apply to the use of such diagrams the diagram by definition is a non-computational way of doing something you couldnt do with ordinary math eg. formal logic. Contradiction and paradox violate the rediculous rules of "logic". Which is what Carrol demonstrated with ABSTRACT propositions. You can use the components of math and logic to make stupid and silly things that show what happens when you try to use non-human thinking and reasoning to computational or mechanical constructions like syllogisms.

Hi Greg,

Well, yes, you can do diagram representation of valuations while disallowing sentences from being simulatenously. But if you do that, then as long as you accept the account of consequence on which A is a consequence of B iff every model of B is a model of A, you can't avoid ECQ. For you end up with the claim that no model models a contradiction, and hence it becomes vacuously true that any model of the premise(s) is a model of the conclusion.

But yes, if you do the same thing as for FDE, but disallow the case where a sentence is both true and false, you end up with a slightly different (non-paraconsistent) logic, I think K3 (Kleene's three-valued logic) which allows for truth-value gaps but not gluts.

cheers

Good morning!

It looks like the FDE (or LP) semantics above don't meet Greg's desideratum from the original post: that we get rid of ECQ without providing a counterexample to it. After all, the diagrams that allow for regions to be both empty and non-empty provide counterexamples to ECQ.

I'd think the closest thing to what you're after is in the early relevance logicians---Entailment vol. 1 and the like. They don't have a semantics for the logic yet (and so counterexamples are out of the question), but they still reject ECQ. Proof-theoretic conceptions of validity seem to be the key, since they don't use counterexamples at all. (Stephen Read's stuff on relevant logic is a nice middle ground here; it might be worth checking out too.)

Also: I don't think the logic you describe in your post is FDE, since A&~A entails A in FDE. But if there's no diagram of A&~A, and that's why A&~A doesn't entail B, presumably it won't entail A either.

I'd think you could get a semantics for the diagram logic just by adapting classical semantics as follows: an argument is valid iff the premises are all true on at least one interpretation, and every interpretation that satisfies the premises satisfies the conclusion. I don't know of a proof-theoretical characterization of that logic, though. (Note that it would fail inferences like Simplification: A&B/A, since B might be ~A.)

Hello Dave,

"Also: I don't think the logic you describe in your post is FDE, since A&~A entails A in FDE. But if there's no diagram of A&~A, and that's why A&~A doesn't entail B, presumably it won't entail A either."

Well, this depends on what you mean by "representing".

Recall, when we switch to representing sentences, we can mark a region empty or non-empty. this means, say, crossing it out, or putting a plus sign in it. You can also do both.

Now, to check if A |- B, you mark A non-empty, follow all the rules (see above), and check if you end up having B unmarked.

So suppose you mark A&~A non-empty. This means you have to mark both A and ~A as non-empty. But marking ~A as non-empty requires also marking A as empty. And marking A as empty requires marking ~A as empty.

So if by representing a premise you mean FIRST following the rules, till you break down truth conditions to markings of atomic formulas, you can represent A&~A using a single diagram with one circle in it, two plus signs, inside and outside the circle, and both regions being simultaneously crossed out. But in such a picture, A is also (among other things) marked non-empty, so A follows from A&~A.

What you're getting at, I guess, is that I suggested thinking about conjunction as intersection, and there's no intersection Of A and ~A in the diagram. If you don't buy into breaking down truth conditions first, you can think of an empty intersection as the line marking the circle itself (of course, crossing it out becomes difficult for technical reasons, etc.). This is an interesting point, and it indicates that if you really think of conjunction as intersection and that you don't want to break down truth conditions first and then mark representations of atomic formulas, you have to admit that a line itself (=the empty set) can also be marked.

As for relevant logics, there are semantics for many of relevant logics. They're relational and employ Routley's star, or three-termed accessibility relation. (cf Mares' "Relevant Logic", ch. 2). And ECQ has counterexamples in those semantics.

Rafael:

Totally. When I said "your post", I meant Greg's original post, where he was looking to fail ECQ without giving counterexamples.

Of course, in the system you give, A&~A entails A. But I think that's just why it doesn't answer to what Greg was after, if I understand his original post right.

Thanks for all the helpful comments, everyone. This thread is long-dead by now, but I just wanted to record two very quick points, in case I ever come back to thinking about this stuff.

1. As many folks correctly noted, simplification(aka 'and'-elimination) fails, since 'A and not-A' does not entail A. But I now realize it's even worse: the rule of reiteration (or whatever you call it) also fails: from

(P1) A

(P2) 'not-A'

you cannot infer A.

2. Though general EFQ is blocked, a more limited form of irrelevant conclusion CAN be inferred from a contradiction: from

'A and not-A'

you can infer

'B and not-B'

since the Venn diagrams of those two sentence-forms are the same.

Yes, it seems long-dead. I'm just cleaning up my to-do list before holidays.:) So...

If you represent premises the way I described in my last comment (i.e, you first follow decomposition rules, then mark truth conditions of atomic sentences, and then assess the truth-value of the conclusion), then your points 1 and 2 don't hold:

ad 1 - just draw a plus sign in the A circle and outside of that circle. This marks A&~A. Then, of course, A is marked, so A follows from A&~A.

ad 2 - say you have two circles in your diagram. Put + in: A-but-not-B

and outside of A and B.

Then, A&~A is marked, even though B&~B isn't (because B isn't).

Having said that, I think I agree that this doesn't satisfy the desideratum of invalidating ECQ without providing a counterexample, if that's what you were after.

Well explained. Keep updating

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