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Deontic logic often contains the axiom

□(p → q) → (□p → □q)

where □ is being used for "it is obligatory that".

This axiom strikes me as odd. It reads "If it is obligatory that p implies q, then if it is obligatory that p, then it is obligatory that q. For example, let

p = X has the ability to save Y's life
q = X saves Y's life

With these definitions, we haves for the premise of the axiom:

A. It is obligatory that if X has the ability to save Y's life, then x saves Y's life.

According to the axiom, A implies B:

B. If it is obligatory that X has the ability to save Y's life, then it is obligatory that X saves Y's life.

Isn't that an odd conclusion? In this case, □p doesn't even make sense, and in general the obligatoriness of a the premise of an implication doesn't seem related to the obligatoriness of the conclusion. From A, shouldn't we instead conclude C?

C. If X has the ability to save Y's life, then it is obligatory that X saves Y's life.

This corresponds not to

□(p → q) → (□p → □q)

but to

□(p → q) → (p → □q)

which strikes me as a much more intuitive axiom. I'm certain that this has been explored by someone, but the SEP article on deontic logic doesn't mention it (although it does discuss problems with this axiom and other solutions). Can anyone else explain why my suggestion doesn't work or give me references to discussions?

David Gudeman
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  • I think this has to do with expressing conditional obligations in SDL: condition p makes q obligatory. Von Wright formalizes it as □(p → q) and Chisholm as p → □q, but both versions lead to problems with Chisholm's paradox. The usual conclusion is that SDL is not suitable for conditional obligations, only for absolute ones, and some revisions are necessary, see [Duc, On a Dilemma of Conditional Obligation](https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=f99b26376cc0ad8508a2548848ea138d52f8e5c1). – Conifold May 22 '23 at 21:07
  • While I sympathize with yer objection (despite the error ye made, it's thought provoking). I suggest ye take a closer look at what exactly ye put in the deontic bag there. Lemme end with a big thank ye! – Agent Smith May 23 '23 at 02:39

2 Answers2

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Neither □(p → q) → (□p → □q) nor □(p → q) → (p → □q) really work. Both run into the problem pointed out by Chisholm with his contrary to duty paradox. The problem is that the simple version of deontic logic does not correctly handle cases where something forbidden is true.

More generally, if you try to represent a conditional obligation as □(p → q) then you run into the problem that □p entails □(¬p → q) for any arbitrary q. If you instead try to represent a conditional obligation as p → □q then ¬p entails p → □q for any arbitrary q. The upshot is that trying to represent conditional obligations using only a unary obligation operator and the material conditional results in conflicting or arbitrary obligations.

There is an analogy here with conditional probabilities. You cannot correctly represent the probability of "if A then B" either as P(A → B) nor as A → P(B). Conditional probabilities are represented using a primitive dyadic function P(B|A) which is not simply a function of the truth values of A and B, nor of the probabilities P(A) and P(B). In similar vein, some versions of deontic logic represent conditional obligations as a primitive dyadic OB(B|A). Even then, there are difficulties handling conflicting obligations, which has led to some to abandon the attempt to develop a logic of absolute obligation in favour of a system of comparative ranking of obligations.

Bumble
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  • Hello, I do not understand your last paragraph, about probabilities. 'The probability of "if A then B"' is literally P(A → B) so I don't know what you mean by "cannot represent". Also, A is an event and P(B) is a number, so "A → P(B)" doesn't make any sense. – Stef May 23 '23 at 07:20
  • @Stef The → here is the material conditional, so A → B is equivalent to ¬A ∨ B. The probability of "if A then B" is given by P(B|A), while the probability of P(A → B) is equal to the probability P(¬A ∨ B). But P(B|A) and P(¬A ∨ B) are completely different functions. For example, if I roll a regular die and A is "the die comes up even", and B is "the die comes up six", the probability of "if A then B" = P(B|A) = 1/3. But the probability P(A → B) = P(¬A ∨ B) = 2/3. The material conditional is systematically wrong when conditionals are uncertain. – Bumble May 23 '23 at 22:35
  • @Stef I am assuming a Bayesian use of probability here, so A and B are not events but propositions. P(A) represents the probability that A is true, and P(B|A) represents the conditional probability that B is true given A is true. Thus A → P(B) would be understood as the probability that B is true, assuming A is true, or equivalently, either A is false or B has the probability P(B). As before, trying to represent conditionals this way gives the wrong result. – Bumble May 23 '23 at 22:35
  • Okay. This might be a vocabulary issue. I just don't understand the words you use. P(B|A) is the conditional probability of B knowing A, not the probability of "if A then B". The probability of "if A then B" is P("if A then B") = P(A → B). It's literally what it says. I agree that "A→B" and "¬A ∨ B" are equivalent but I don't know why that would be an issue. Also, I still have no idea what you mean by A→P(B). P(B) is a number. If A="the die comes even" and B="the die comes up 6" then "A→P(B)"="if the die comes even then 16.67%"? – Stef May 24 '23 at 07:18
  • Apologies if I sound obtuse or aggressive. That really wasn't my intention. But I genuinely don't understand your words or notations. – Stef May 24 '23 at 07:19
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    @Stef The conditional probability P(B|A) is the probability of B supposing A to be true. We do not have to know that A is true in order to put a value to P(B|A). In typical cases, the probability of "if A then B" is given by P(B|A) and not by P(¬A ∨ B). What is the probability that if England reach the final of the next World Cup they will win? Fair to middling. What is the probability that England don't get to the final or they win? Extremely high, since they probably won't get to the final. – Bumble May 24 '23 at 21:30
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    Similarly with the example of the die. What is the probability that if it comes up even it is a six? One third. What is the probability that it comes up not even or a six? Two thirds. These are not just one-off examples. There have been many published studies by cognitive psychologists in which people are presented with some information and asked, What is the probability of "if A then B" in this situation?. The answer is nearly always P(B|A) and not P(¬A ∨ B). – Bumble May 24 '23 at 21:31
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    The probability of "if A then B" and the probability P(A → B) are different things, and hence "if A then B" and A → B are different things. Handling uncertainty is one of the many cases in which the material conditional fails to correctly capture the meaning of "if A then B". I listed some others in a recent answer to another question. https://philosophy.stackexchange.com/questions/99353/what-are-the-arguments-of-philosophers-against-the-reasoning-which-justifies-the – Bumble May 24 '23 at 21:31
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    As to A → P(B) ...I only mentioned it because I have seen somebody try to defend it. Since we both agree it doesn't work, there's no point saying anything more about it. There's no need to apologise by the way; I enjoy discussing logic and conditionals and I'm always happy to be challenged. – Bumble May 24 '23 at 21:31
  • If someone asked me "What is the probability that if it comes up even it is a six?", I would ask them to rephrase the question; and if they insist that I answer it literally, I would say the probability is 0, because "if it comes up even it is a six" is always false, so the probability of "if it comes up even it is a six" is 0. And if someone asked me "What is the probability that if England reach the final of the next World Cup they will win?", I would indeed interpret it as P(England win | England reaches final), but I would also note that the grammar of the sentence was dubious. – Stef May 30 '23 at 16:05
  • On its own as a full sentence, I would argue that "If A then B." is exactly the same as "A → B". However, as part of a longer sentence, "if A then B" can be interpreted differently if it's not surrounded in quotes, because of how the sentence is parsed; for instance, I would interpret "What is the probability that if England reach the final of the next World Cup they will win?" as "Assuming that England reaches the final, what is the probability that they win?" rather than as "What is the probability that [if England reach the final of the next World Cup, then they will win]?" – Stef May 30 '23 at 16:08
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    @Stef I don't see why "if it comes up even it is a six" is false. Nor do I see any problem with the grammar of "the probability that if England reach the final... they will win". But grammar aside, the point remains that were it the case that A → B gives the correct truth conditions of "if A then B" then the probability of "if A then B" would be the probability of P(A → B) which is also P(¬A ∨ B). And it just isn't. The fact that you have to parse the conditional as "the probability of B assuming A" demonstrates that "if A then B" is not a proposition with the truth conditions of ¬A ∨ B. – Bumble May 31 '23 at 03:43
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    There is in fact one theory of conditionals, called the suppositional theory, under which conditionals are not propositions and do not have truth values. On this view, conditionals are always understood as "B under the supposition A", so what is thought of as the truth value of "if A then B" is really the truth value of B within the scope of the supposition A. Likewise for the probability of B. You can still derive the logic of classical connectives from this theory. – Bumble May 31 '23 at 03:43
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In the FOL with which SDL is built, p → q can be converted into ¬p ∨ q as well as ¬(p ∧ ¬q) (I wasn't clearly aware of this until a few minutes ago, from reading Bumble's answer to this PhilosophySE question). So:

  1. OB(p → q) = OB(¬p ∨ q)
  2. OB(p → q) = OB¬(p ∧ ¬q)

Then:

  1. OB¬p ∨ OBq
  2. However, I don't know how to phrase the repurposing of (2): OBp ∧ OB¬¬q?

I don't think you're wrong about the facts-on-the-ground, though. And I'm not even sure that saying a whole conditional is obligatory is always intelligible in the first place. It seems similar to saying that it would be obligatory for something to be possible simpliciter, whereas I suspect it is more realistic only to say that it might be obligatory for something to be accessibly possible (true in an accessible world, not just generically possible world).

So now I also think that a proper interpretation of an OB-conditional-as-a-whole might depend on combining temporal and deontic logic. I haven't read through it yet to see if they bring up your kind of concerns, but this sciencedirect.com entry appears to be about such a combination.

Kristian Berry
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    I don't really follow your transformations, but the point that material implication is equivalent to a disjunction pretty much answers the question. Converted to a disjunction it's impossible to justify treating the two sides of the disjunction differently. – David Gudeman May 22 '23 at 23:06