SDL Containment Proof

Recall SDL:

A1: All tautologous wffs of the language	(TAUT)
A2: OB(p → q) → (OBp → OBq)	(OB-K)
A3: OBp → ~OB~p	(OB-NC)
R1: If p and p → q then q	(MP)
R2: If p then OBp	(OB-NEC)

We have already shown OB-NC is derivable in Kd above, and TAUT and MP are given, since they hold for all formulas of Kd. So we need only derive OB-K and OB-NEC of SDL, which we will do in reverse order. Note that RM, if r → s, then □r → □s), is derivable in Kd, and so we rely on it in the second proof.^[1]

Show: If p then OBp. (OB-NEC)
Proof: Assume p. It follows by PC that d → p. So by NEC for □, we get □(d → p), that is, OBp.

Show: OB(p → q) → (OBp → OBq). (K of SDL)
Proof: Assume OB(p → q) and OBp. From PC alone, (d → (p → q)) → [(d → p) → (d → q)]. So by RM for □, we have □(d → (p → q)) → □[(d → p) → (d → q)]. But the antecedent of this is just, OB(p → q) in disguise, which is our first assumption. So we have □[(d → p) → (d → q)] by MP. Applying K for □ to this, we get □(d → p) → □(d → q). But the antecedent to this is just our second assumption, OBp. So by MP, we get □(d → q), that is, OBq.

Metatheorem: SDL is derivable in Kd.

Note that showing that the pure deontic fragment of Kd contains no more than SDL is a more complex matter. The proof relies on already having semantic metatheorems available. An excellent source for this is Åqvist 2002 [1984].^[2]

Return to Deontic Logic.