Counterexamples to Modus Ponens?

August 19, 2008

Moritz has a very interesting post on McGee’s counterexample to modus ponens. One thing that came up in the comments was how things looked on the restrictor analysis of conditionals. On this view ‘if’ is not to be thought of syntactically as a connective, but rather as a device for restricting modals. The basic idea is, given a modal, O, and antecedent and consequent, p and q, ‘if’ allows us to restrict the domain of O to p-worlds. Roughly, ‘if p, Oq’ means that q is true in all the best O-accessible p-worlds. In this note I want to argue that the restrictor view will have to admit exceptions to modus ponens even for simple (non nested) conditionals.

Moritz points out that, since we are no longer dealing with a connective, it is not clear what counts as an instance of modus ponens. One reaction, then, might be to simply say that there are no instances of modus ponens for the English conditional; the rule is only applicable to connectives, so we should reserve the rule for formal connectives that behave in the right way syntactically. I think this is a little extreme – after all, modus ponens presumably originated from a rule of inference for natural language conditionals, and we seem to be able to recognise instances of it when we them in natural language reasoning.

On the syntactic understanding of modus ponens for connectives the rule is something like the following: from p and p \rhd q infer q (where \rhd is syntactically a connective.) Thus we can say, for example, that modus ponens holds for the intuitionist conditional, the Stalnaker conditional, even for classical conjunction, but not for classical disjunction, the reverse conditional (\leftarrow) and so on. But even here we’re not home and dry. We still need to be able to distinguish the antecedent from the consequent syntactically. Does modus ponens hold for the connective \frac{p}{q} defined to have the same truth conditions as p \rightarrow q? Depending on which counts as the antecedent, we get the material conditional or the material reverse conditional – MP holds for the former but not latter.

In short, I think an explicit syntactic characterisation of MP is not the correct way to go. For now I think we can just settle for a rule of thumb: if we can recognise it as in instance of modus ponens in English, then its underlying logical form forms the basis for an admissible syntactic characterisation for modus ponens. This allows us to characterise modus ponens even for syntactic items that aren’t connectives. Indeed, if English conditionals really do have the logical form the restrictor analysis predicts, then the following schemas are plausible syntactic characterisations of MP. (Notation: if \Box is an operator, \mbox{[if }p: \Box ] is the new operator formed by restricting by p.)

  1. From p and \mbox{[if }p: \Box ]q, infer q
  2. From p and \mbox{[if }p: \Box ]q, infer \Box q

Now I’m not going to defend the claim that both of these form genuine rules that capture modus ponens for the English conditional. Rather, all I want is that there are no other candidate rules to fill this role which are remotely plausible. This can be argued for by brute force. We have no difficulty identifying the antecedent and the conditional, thus premisses are definitely as represented as above. Combinatorially, there are only four candidates for the conclusion that we can construct from the available items: p, q, \Box p, and \Box q. Clearly we can eliminate p and \Box p.

Let us start with (1). Since the type of ‘if’ is a relation between two propositions and an operator, (1) is only valid if no matter how we reinterpret p, q and \Box the conclusion is true whenever the premises are. A related question, which will determine the answer above one, is to ask for what interpretations of \Box is the above valid (i.e. fixing the interpretation of \Box and varying only p and q.) It turns out (1) is valid relative to an interpretation of \Box iff it has a reflexive accessibility relation. Let’s consider just the direction of the biconditional that invalidates (1): it is perfectly possible for p to be true at a world w, and for q to be true at all the accessible p-worlds, and q false at w so long as w isn’t among the accessible p-worlds. Since w is a p-world, it must be inaccessible – i.e. this happens when the accessibility relation isn’t reflexive.

To illustrate, consider an example where the restricted modal isn’t reflexive. The easiest cases to consider are doxastic and deontic modals.

  • John’s a murderer
  • If John’s a murderer, he ought to be in jail
  • Therefore, John is in jail

This is a counterexample to (1), there is not enough evidence to put John away. You might object – surely this was not a plausible characterisation of modus ponens in the first place. What we should rather conclude from the first two premises is that John ought to be in jail. This is essentially an appeal to (2).

Unfortunately (2) admits counterexamples as well. Suppose there are accessible ~p worlds, where q is false, but q is true at all the accessible p worlds, and that p is true at the actual world. Certainly p is true, and \mbox{[if }p: \Box ]q is true because q is true in all the accessible p-worlds. However, \Box q is false, since there are accessible p-worlds where q is false.

Examples are difficult to conjure up, since modals are often sensitive to the conversational background. Whenever I assert p, I essentially restrict the \Box-accessible worlds to p-worlds. Thus asserting p and then \mbox{[if }p: \Box ]q invariably places us in a context according to which \Box q is true. Let us try anyway. Suppose that for all we know, Jones isn’t in his office. So: it is not the case that Jones must be in his office. Be that as it may, if it is 3.00pm, Jones must be in his office, because we know his lunch break ends at 2.00pm. Suppose further that unbeknownst to us, it is be 3.00pm. That sounds like a consistent story right? But now if I assert them in the following order:

  • It’s 3.00pm
  • If it’s 3.00pm, then Jones must be in his office
  • It’s not the case that Jones must be in his office

the trio above sounds inconsistent. But strictly speaking they’re all true, provided we keep the same context throughout.

What is interesting is that the McGee counterexamples, on a plausible syntactic analysis, form more convincing counterexamples to modus ponens on characterisation (2). For on a plausible syntactic analysis of nested conditionals, ‘if p, if q, then must r‘ the inside conditional ‘if q, then must r‘ has the form s = \mbox{[if }q: \Box ]r. Since s has the form operator:proposition, ‘if p, s’ is naturally read with ‘if p’ restricting the compound operator, rather than some covert modal. Thus the embedded conditional gets the form \mbox{[if }p: \mbox{[if }q: \Box ]]r. Essentially we are doubly restricting \Box. The resulting McGee example is not an instance of modus ponens on characterisation (1). However it does represent a failure for the rule (2), where the modal is the compound operator \mbox{[if }q: \Box ].


  1. Hi Andrew,

    sorry that it took me so long to read this entry. After Krakow we had another workshop (the “phloxshop”) here in Berlin.

    I am still a bit troubled by how best to set up the debate about modus ponens within the restrictor’s framework. I need to think more about this.

    Perhaps there is a way to strengthen your second exammple. Once we enter the realm of epistemic modals etc, it is not clear whether asking what is assertable in which contexts is a very safe guide to validity. For instance, in a context in which you assert “p”, you may also be committed to assert “Must p”. But that shouldn’t force us to think that “Must p” follows from “p”. Rather, we should focus on the fact that assertion is governed by a certain norm, perhaps knowledge. Now, asserting “p” without acknowledging that the norm governing this assertion is in place is pragmatically inappropriate. So, asserting “p” and refraining from asserting “I know p” seems to be pragmatically somewhat strange. Nevertheless, no way that the latter follows from the former.

    Now, I think that your example sounds not as good as you would like it to be for precisely this reason. But perhaps we can look at another feature of this inference in order to see that it is not valid. It seems that we can have high credence in the first premise, high credence or even certainty in the second premise, but low credence in the conclusion. Usually, such a complex of epistemic attitudes is not rationally permitted with respect to a valid inference. Doesn’t this show what you want?

  2. Hey Moritz,

    Hope you had a good time in Krakow. I was planning on going, but was too disorganised :-s.

    Thanks for your comments! I like the method of using probabilistic consequence to get a handle on inferences where standard methods are left wanting in the way you describe. However, I think that this consequence relation differs from the relation we think of as ‘logical consequence’ in some crucial respects, that make its application here slightly spurious.

    In fact – I think the probabilistic notion is actually *extensionally* inadequate in that it classifies some inferences as valid which aren’t (in the strictly logical sense.) The example I have in mind trades on Moore like paradoxes: ‘p but I have a low credence in p’. In particular the following inference is valid with respect to all rational credence functions:

    * (p \wedge Cr(p)=0) \models (q \wedge \neg q)

    (The argument trades on the S4 principle for rational certainty – I briefly spelled it out here if you’re interested.)

    Now the reason I think this is relevant here is because the counterexample above trades on doxastic and epistemic attitudes in the premises and conclusion, and the counterexample to modus ponens presented trades on epistemic modals.

    Another potentially damaging principle relevant to this context, which governs the interaction between credences and knowledge, is the principle that if you believe something, then you believe that you know it. If you accept the corresponding principle for full credence instead of belief that makes the following valid:

    * p \models Kp

    Even if you found the Moore like paradox above an acceptable inferenc, this one is clearly an awful inference.

  3. Sorry, I think I was a bit too quick on some of that. In particular, there are two notions of probabilistic inference that are relevant, which makes a difference to inferences I gave.

    If we think of validity as preservation of certainty, we get

    * (p \wedge \neg Cp) \models  (q \wedge \neg q)
    * p \models Kp

    Assuming the principles (Cp \rightarrow CCp) and (Cp \rightarrow CKp) respectively. (Here Cp means full credence in p.) This consequence relation violates classical inferences rules though, for example the deduction theorem:

    * If p \models Cp then \models (p \rightarrow Cp)

    To see how the consequent fails, it seems perfectly rational to have a credence of 1/2 in p and 0 in Cp. (I think most of the Williamson arguments against validity=preservation of supertruth carry over here.)

    The other relevant consequence relation is where the sum of the uncertainties of the premises must be greater-than-or-equal to the uncertainty of the conclusion. This gets us back classical inference rules, like DT. I assume you had this one in mind, while I was assuming the other one in my post. The Moore like inference I stated in the first post remains valid on this consequence relation given the S4 axiom for certainty (and it was more than I needed for validity on the first consequence relation.) For the second inference you get it if you assume that your credence in p should be equal to your credence that you know p: Cr(p)=Cr(Kp). I find the principles appealed to quite appealing – but maybe you disagree.

  4. Hi Andrew,

    yes, I had the latter conception of probabilistic validity (a la Adams) in mind.

    I don’t know about the S4 axiom for certainty. But the second principle you mention seems to be false (if I understand it correctly). Consider a lottery case. My credence in `My ticket will lose’ is high, but strictly less than 1. However, I am certain that I do not know that my ticket will lose. This seems to be a rational epistemic state. But it would violate the principle that my credence in “p” equals my credence in “I know that p”.

  5. What about Williamson’s counterexamples to the KK-principle? Don’t they carry over to certainty?

  6. Hmm. In the lottery case you described I’d be more inclined to say that I’m pretty sure I know I’m not going to win, even though there’s a small chance I’m wrong. Maybe I just have deviant intuitions.

    Actually I’ve never been quite convinced about the failure of KK. But will the Williamson argument carry over to certainty? I’m not sure – if you’re certain that p_k, then you’re credence in p_n where n!=k is 0 by finite additivity. Doesn’t that just mean the margin for error principle fails? You might also think you should have fuzzy credences in propositions like in the Williamson example, in which case I don’t know how the argument would run.

    Eitherway, you don’t actually need anything as strong as S4 for certainty for the result I mentioned. You can do it with: C¬Cp->¬Cp, which is strictly weaker. Another route would be that there can’t be any rational credence function that gives full credence to p, and full credence that p has been assigned zero credence. The counterexamples to KK rely on small differences between the credences and the credences in the credences, but here you have the widest difference possible.

  7. Guess I need to think more about higher-order credences…

    Concerning the lottery case: there is an even simpler example. Just think of a typical heads-tails situation. Your credence in heads is 1/2. But your credence in that you know that the coin comes up heads is 0, since you are certain that you cannot know whether heads or tails. Do your intuitions still differ?

  8. Yeah, I share the intuition with that example.

    The crux of the problem I was worried about is that there are various constraints on rational credences that ensure some valid probabilistic inferences are logically invalid. For example there are surely *some* constraints regarding higher order credences. Similarly, you would have thought there are some connections between credences and knowledge (e.g. if you know p, you should have a high credence in p or if you have a high credence that you know p, you should have a high credence in p, and you should have a high credence that you have a high credence in p. etc etc…) If you accept some of these connections you can construct more problems along the same lines.

  9. That may be right …

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    • Going to put this arltice to good use now.

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