# Justification for axiom OB-RE in deontic logic

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
• 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
• – Eliran Dec 8 '18 at 1:08
• Can someone tell me what the RE here stands for? I came to this question from the Stanford site looking for an answer. – weirdalsuperfan Jul 19 '19 at 11:41
• @weirdalsuperfan Good question. I don't know the answer, but I know that axiom RE exists in other modal logics as well. See here and here (page 4) for instance. – user76284 Jul 19 '19 at 17:34