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):
- O(P ⊃ Q)
- ∴ (OP ⊃ OQ)
- Assume ~(OP ⊃ OQ) {assume the conclusion in 2 is false}
- ∴ (OP · ~OQ) {not-if, from 3}
- ∴ ~OQ {Conjunction elimination, from 4}
- ∴ OP {Conjunction elimination, from 4}
- ∴ R~Q {Reverse squiggle, from 5}
- D ∴ ~Q {Drop R and introduce a deontic world D, from 7}
- D ∴ P {Drop O in deontic world D, from 6}
- D ∴ (P ⊃ Q) {Drop o in deontic world D, from 1}
- D ∴ Q {Modus Ponens, from 9 and 10}
- 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