modal logic

It is usually considered that deontic logic is better captured by system D in modal logic. but I don't see exactly why, as deontic logic does not normally accept axiom T. For example, this statement ⊨D [▢P ∧ ▢(∼P ∨ Q)] → ◇Q is a logical truth in D, but not so clearly true in deontic logic, I think:

PROOF
We must show that V([▢P ∧ ▢(∼P ∨ Q)] → ◇Q, w) = 1 for the valuation V for an arbitrarily chosen D-model〈W, R, I〉and world w in that model.

(i) Assume for reductio that V([▢P ∧ ▢(∼P ∨ Q)] → ◇Q, w) = 0
(ii) So V(▢P ∧ ▢(∼P ∨ Q), w) = 1, and …
(iii) … V(◇Q, w) = 0.
(iv) From (ii), and by the derived truth conditions for ∧, V(▢P, w) = 1, and so by the truth conditions for the ▢, P is true in every world accessible from w. Since our model is a D-model, this means that the accessibility relation R is serial, that is to say, for every u ∈ W, there is some v ∈ W, such that Ruv. So in this case, there is at least some world, call it v, such that Rwv, and therefore V(P, v) = 1.
(v) From (ii), and again by the truth conditions for ∧, V(▢(∼P ∨ Q), w) = 1. So again by the truth conditions for ▢, (∼P ∨ Q) is true in every possible world accessible from w. So in particular, V(∼P ∨ Q, v) = 1.
(vi) For (v), and the truth conditions for ∨, V(∼P, v) = 1 or V(Q, v). By the truth conditions for ∼, V(P, v) = 0, so since for (iv), we know that V(P, v) = 1, it must be the case that V(Q, v) = 1.
(vii) Given (iii), Q is false at every possible world accessible from w, so in particular it must be the case that V(Q, v) = 0. This contradicts (vi).
(viii) Therefore, by reductio, we have proved V([▢P ∧ ▢(∼P ∨ Q)] → ◇Q, w) = 1.
But sometimes I do miss.