# How to get the refutation of (OP ⊃ OQ) ∴ O(P ⊃ Q) in Deontic Logic

In Deontic Logic, one could easily infer `"If it is obligatory that P, then it is Obligatory that Q"`, from `"It is obligatory that if P then Q"`

O(P ⊃ Q) ∴ (OP ⊃ OQ)

Where the is an implication (if ... then).

Consider the following proof (D stands for a Deontic World, R stands for Permissible, and O stands for Obligatory):

1. O(P ⊃ Q)
2. ∴ (OP ⊃ OQ)
3. Assume ~(OP ⊃ OQ) {assume the conclusion in 2 is false}
4. ∴ (OP · ~OQ) {not-if, from 3}
5. ∴ ~OQ {Conjunction elimination, from 4}
6. ∴ OP {Conjunction elimination, from 4}
7. ∴ R~Q {Reverse squiggle, from 5}
8. D ∴ ~Q {Drop R and introduce a deontic world D, from 7}
9. D ∴ P {Drop O in deontic world D, from 6}
10. D ∴ (P ⊃ Q) {Drop o in deontic world D, from 1}
11. D ∴ Q {Modus Ponens, from 9 and 10}
12. Line 11 contradicts 8, therefore the assumption must be false, we apply Reductio Ad Absurdum to conclude that ∴ (OP ⊃ OQ)

As you can see, we can prove that `O(P ⊃ Q)∴(OP ⊃ OQ)` , but how can we prove or refute the inverse :

`(OP ⊃ OQ) ∴ O(P ⊃ Q)`

I spent 2 hours now trying to obtain a refutation

Edit

Here, reverse squiggle means to go from `It is not obligatory that X`, to `It is permissible that not-X`

``````~OX ⇔ R~X
``````

If it is not the case that you ought to do X, then it is permissible to do Not-X

I don't doubt the conclusion, but your attempted proof appears to be circular. In your answer, you assume that 1 is true and 2 is false and then reach the conclusion that 2 is false.

Suppose for the moment we drop the modality and use a purely extensional version of the argument. Then we would be asking, does 1E entail 2E:

1E. (∀x)Fx ⊃ (Ay)Gy

2E. (∀x)( Fx ⊃ Gx )

Clearly the answer to that is no. If some things are F and some not, then 1E is true because it has a false antecedent, but 2E may be false, because it is not necessarily the case that all those things that are F are also G. For example, it is true that "everything is a duck" materially implies "everything is a reptile", because not everything is a duck so the antecedent is false, but it does not follow that "everything that is a duck is a reptile".

In a modal logic, the same kind of reasoning applies, except you are quantifying over worlds rather than things. If P is a contingent circumstance then OP is false and hence (OP ⊃ OQ) is true. But it does not follow that O(P ⊃ Q) since there may be some world in which P is true and Q does not hold.

There is however a much more important problem. Deontic conditionals cannot correctly be represented as strict implications, that is to say, as material implications within the scope of a deontic operator. Material implication is just a special case of a conditional, and it does not apply to conditional obligations. Attempting to represent conditional obligations as material implications leads to several paradoxes, including Chisolm's contrary to duty paradox. There are different ways to express conditional obligations, including treating them as primitive dyadic functions, or using non-classical logic. The SEP article on Deontic Logic has some useful references.

• I reached the conclusion that 2 is true applying a refutation, because there is a world where ~O(P > Q) is true, so the argument has to be invalid, because there exist cases where (OP > OQ) does not lead you to O(P > Q) necessarily. I am not saying that it follows necessarily that ~(OP > OQ) , I am saying that since there is a deontic world where ~Q and P is the case, and where the (OP > OQ) in the premise holds true , then the argument is invalid. – SmootQ Jan 7 at 12:43
• I didn't reach the conclusion that 2 is false, I reached the conclusion that 2 is not necessarily true, as there is a combination of an actual world and deontic world where it is false although the premise 1 is true. – SmootQ Jan 7 at 12:45
• "I reached the conclusion that 2 is true applying a refutation", I mean conclusion that 2 does not follow. Typing mistake. – SmootQ Jan 7 at 12:51
• Consider this example of a refutation : (if A then B), B , therefore A .. Let us assume the conclusion is false ~A , then we cannot do anything else to obtain a contradiction caused by our assumption, we got a refutation, in which there is a world where B is true, and A is false, which means that "A" does not follow from the 2 premises since there is already another possibility where ~A , (the assumption). This does not mean that A is false, it means that the conclusion A does not follow from the premises, and this is what I did in my refutation, albeit more complicated. – SmootQ Jan 7 at 20:53

I found a way out, (but I am not very sure : fingers crossed ):

1. (OP ⊃ OQ)
2. ∴ O(P ⊃ Q)
3. assume 1 ~O(P ⊃ Q) {assume the conclusion 2 is false}
4. R~(P ⊃ Q) {reverse squiggle, from 3}
5. R(P · ~Q) {not-iff, from 4}
6. D ∴ (P · ~Q) {drop R and introduce a deontic world D, from 5}
7. D ∴ P {Conjunction elimination, from 6}
8. D ∴ ~Q {Conjunction elimination, from 6}
9. assume 2 OQ {we assume the consequent in line 1}
10. D ∴ Q {We drop O in deontic World D, from 9}
11. from 9, 10 contradicts 8 , which means that assumption in line 9 is false, therefore ∴ ~OQ
12. ∴ ~OP {Modus Tollens, from 11 and 1}

At this point (line 12), we cannot do anything useful, and we can clearly see that we have a Deontic World D where P (line 7) and ~Q (line 8) , and we have an actual world where ~OQ (line 11) and ~OP (line 12)

It seems like this is a refutation (I am not very sure here), and this refutation is as such :

• World D : P, ~Q
• Actual World : ~OP, ~OQ

In the Deontic world D, (P ⊃ Q) evaluates to false, which means that it is not the case that (P ⊃ Q) is Obligatory

Therefore : O(P ⊃ Q) is false

And we can clearly see how (~OQ ⊃ ~OP) evaluates to true in the actual world, therefore : (OP ⊃ OQ) is true).

So, we have a case where the premise is true and the conclusion is false

1. (OP ⊃ OQ) = True
2. ∴ O(P ⊃ Q) = False

Which means that this reasoning is invalid and inconsistent :

"If it is obligatory that P then it is Obligatory that Q" implies that "It is obligatory that if P then Q" .

While this reasoning is consistent and valid (as we saw in the first proof) :

"It is obligatory that if P then Q" implies that "If it is obligatory that P then it is Obligatory that Q" .

So you could directly infer a conditional obligation, from the obligation of the antecedent and the consequent, but not the other way around.

I hope this helps.