Let OB p denote "p is obligatory". Axiom OB-RE is (p ↔ q) → (OB p ↔ OB q). This axiom seems false to me (under the interpretation of obligation). For example, let p denote "don't lie to me" and q denote "hurt me". Suppose p Λ q, that is, you're not lying to me but you are hurting me. Then it is true that p ↔ q, since ⊤ ↔ ⊤. Suppose furthermore that OB p, that is, you're obligated not to lie to me. Then OB-RE would imply OB q, that is, you're obligated to hurt me. This clearly doesn't seem right. Am I interpreting this axiom incorrectly?

  • No. The entry states "If p ↔ q is a theorem". You left that crucial part out. p ↔ q is not a theorem in your example. – Eliran Dec 8 '18 at 0:54
  • @Eliran How is p ↔ q not a theorem? Since p is true, q → p (any material conditional with a true consequent is true). Similarly since q is true, p → q. Therefore, p ↔ q. – user76284 Dec 8 '18 at 1:04
  • A theorem in logic means that it can be derived without assumptions. In other words (and given completeness), p ↔ q is a theorem when p ↔ q is a tautology. That doesn't hold in your case. – Eliran Dec 8 '18 at 1:07

OB-RE: If p ↔ q is a theorem, then so is OBp ↔ OBq.

In other words, if p and q have the same truth value in each world, and each model, then they're basically the same proposition, and so one is obligatory iff the other is.

This holds in classical systems and almost characterizes them. Classical systems correspond to minimal models, those where what is necessary in a world x is just some collection of sets of world. Here a sets of worlds corresponds to a proposition p which holds in exactly those worlds.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.