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Supplement to Deontic Logic

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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 pandp→qthenq(MP) R2: If pthenOBp( OB-NEC)

We have already shown **OB**-NC is derivable in
K*d* above, and TAUT and MP are given, since they hold for
*all* formulas of K*d*. 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
K*d*, and so we rely on it in the second
proof.^{[1]}

Show: IfpthenOBp. (OB-NEC)

Proof: Assumep. It follows by PC thatd→p. So by NEC for □, we get □(d→p), that is,OBp.

Show:OB(p→q) → (OBp→OBq). (K of SDL)

Proof: AssumeOB(p→q) andOBp. 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 K*d*.

Note that showing that the pure deontic fragment of K*d*
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]}

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Paul McNamara mcnamara.p@comcast.net |

Stanford Encyclopedia of Philosophy