Be Nice! How Simple Imperatives Simplify Imperative Logic

Abstract

In a series of articles, P. Vranas recently proposed a new imperative logic. The strong and weak inferences of this logic are motivated by an appeal to a strong and weak ‘support by reasons’ that transfers from the premisses of an argument to its conclusion. They also combine nonmonotonic and monotonic reasoning patterns. I show that for any moral agent, Vranas’s proposal can be simplified enormously.

Appendix

I prove that my deductions equivalently characterize Vranas’s logic.

Theorem 1 (Equivalence of inferences and deductions)

For any finite set of imperatives {I ₁,...,I _n , I}

I ₁,..., I _n ∣∽ _w I iff I can be deduced from I ₁,..., I _n ,

I ₁,..., I _n ∣∽ _s I iff I can be strongly deduced from I ₁,.., I _n .

Proof

Left-to-right: Multi-premiss: If C ₁ ⇒!A ₁,..., C _n ⇒!A _n ∣∽_∗ D⇒!B, repeatedly use (DisCo) to combine all premisses and then use (CExt) to obtain Conj [C ₁ ⇒ !A ₁,...,C _n ⇒!A _n ]. Continue as for single-premiss inferences with Conj [C ₁ ⇒!A ₁,..., C _n ⇒!A _n ] ∣∽*_∗ D ⇒ !B. Single-premiss: Assume C ⇒ !A ∣∽_∗ D ⇒ !B. If ∗ = s and ⊢ C ∧ ¬A, use (LLE) to replace C by ⊤ and (CExt) to replace A by ⊥. Use (EFQ) to obtain D ⇒!B. Else it must be that ⊢ D → C. Use (SA) to obtain (C ∧ D) ⇒!A and (LLE) to obtain D ⇒ !A. Also it must be that ⊢ D → (A ↔ B). Use (CExt) to obtain D ⇒ !B. If ∗ = w then again it must be that ⊢ D → C, and we obtain D ⇒!A as before. Also, it must be that ⊢ D → (A → B), hence ⊢ D → (A ↔ (A ∧ B)). Use (CExt) to obtain D ⇒!(A ∧ B), (WC) to obtain D ⇒!((A ∧ B) ∨ B), and again (CExt) to obtain D ⇒!B.

Right-to-left: We prove by induction on the length of a sequence that any imperative I in it can be inferred from the premisses above it (where by these I mean either I itself, if it is a premiss, or otherwise what would be the premisses if the sequence were to end with I). Then the conclusion is also inferrable from the premisses of the sequence. Induction basis: If I is a premiss, then we have I ∣∽_∗ I in Vranas’s logic. Induction hypothesis: All heads above a rule can be inferred from the premisses above themselves. Induction step: We need to prove that if the induction hypothesis holds then the tail of a rule application can also be inferred from the premisses above it. The steps for (EFQ), (CExt), (LLE), (SA) and (WC) (for weak inferences) are easy and left to the reader. (DisCo) is a little more tricky since it has two heads. Let

• E ⇒!F and G ⇒!H be the heads of (DisCo),

• C ₁ ⇒!A ₁,…, C _n ⇒!A _n be the premisses above E ⇒!F,

• D ₁ ⇒!B ₁,…, D _n ⇒!B _n be the premisses above G ⇒!H.

By the induction hypothesis,

$$\begin{array}{@{}rcl@{}} C_{1}\Rightarrow!A_{1},...,C_{n}\Rightarrow!A_{n}\mid\thicksim_* E\Rightarrow!F, \\ D_{1}\Rightarrow!B_{1},...,D_{n}\Rightarrow!B_{n}\mid\thicksim_* G\Rightarrow!H, \end{array} $$

which by Vranas’s definition (steps 2 and 3) holds iff

$$\begin{array}{@{}rcl@{}} \mathbf{Conj}\,[C_{1}\Rightarrow!A_{1},...,C_{n}\Rightarrow!A_{n}\,]\mid\thicksim_* E\Rightarrow!F, \\ \mathbf{Conj}\,[D_{1}\Rightarrow!B_{1},...,D_{n}\Rightarrow!B_{n}\,]\mid\thicksim_* G\Rightarrow!H. \end{array} $$

Let C ⇒!A be the first and D ⇒!B be the second conjunction. By step 1 in Vranas’s definition, we have for strong inferences:

• (1) ⊢ E → C and ⊢ E → (A↔F), or ⊢ C ∧ ¬ A

• (2) ⊢ G → D and ⊢ G → (B↔H), or ⊢ D ∧ ¬ B

By the same step we have for weak inferences:

• (3) ⊢ E → C and ⊢ E → (A→F)

• (4) ⊢ G → D and ⊢ G → (B→H)

For the tail I = (E ∨ G) ⇒ !((E → F) ∧ (G → H)) of (DisCo) we need:

$$C_{1}\Rightarrow!A_{1},...,C_{n}\Rightarrow!A_{n}, D_{1}\Rightarrow!B_{1},...,D_{n}\Rightarrow!B_{n}\mid\thicksim_* I,$$

which according to Vranas’s definitions 2 and 3 holds

$$\begin{array}{@{}lccc@{}}\text{iff} {} \qquad\qquad \qquad\qquad\qquad \textbf{Conj}\,[C\Rightarrow!A, D\Rightarrow!B\,]\mid\thicksim_* I, \\ \text{iff} \qquad\qquad (C\vee D)\Rightarrow!((C\rightarrow A)\wedge (D\rightarrow B))\mid\thicksim_* I. \end{array} $$

Consider weak inferences first. First, we must prove that ⊢ (E ∨ G) → (C ∨ D), which is immediate since from (3) and (4) we have ⊢ E → E and ⊢ G → D. Secondly we must prove that

$$\vdash (E\vee G)\rightarrow(((C\rightarrow A)\wedge (D\rightarrow B))\rightarrow ((E\rightarrow F)\wedge(G\rightarrow H))). $$

From (3) we have ⊢ E → C and ⊢ E → (A → F), so also

$$\vdash (C\rightarrow A)\rightarrow (E\rightarrow F). $$

Likewise from (4) we have ⊢ G → D and ⊢ G → (B → H), so also

$$\vdash (D\rightarrow B)\rightarrow (G\rightarrow H). $$

So $\vdash ((C\rightarrow A)\wedge (D\rightarrow B))\rightarrow ((E\rightarrow F)\wedge (G\rightarrow H))$ and we are finished.—For strong inferences, we must prove that either

• (i) $\vdash (C\vee D)\wedge \lnot ((C\rightarrow A)\wedge (D\rightarrow B))$, or

• (ii) $\vdash (E\vee G)\rightarrow (C\vee D)$ and

• (iii) $\vdash (E\vee G)\rightarrow (((C\rightarrow A)\wedge (D\rightarrow B))\leftrightarrow ((E\rightarrow F)\wedge (G\rightarrow H)))$.

Case A: Suppose either ⊢ C ∧ ¬ C or ⊢ D ∧ ¬B. In either case (i) is true, and we are finished.

Case B: Suppose ⊯ _PL C ∧¬A and ⊯ _PL D ∧¬B. This has two consequences: First, it excludes that (EFQ) is applied above a head: Suppose it is applied above C ⇒ A. By the induction hypothesis there are $\{C_{i_{1}}\Rightarrow !A_{i_{1}},...,C_{i_{m}}\Rightarrow !A_{i_{m}}\}\subseteq \{C_{1}\Rightarrow !A_{1},...,C_{n}\Rightarrow !A_{n}\}$ such that $C_{i_{1}}\Rightarrow !A_{i_{1}},...,C_{i_{m}}\Rightarrow !A_{i_{m}}\mid \hspace {-2pt}\thicksim _{s} !\bot$. So $\textbf {Conj}\,[C_{i_{1}}\Rightarrow !A_{i_{1}},...,C_{i_{m}}\Rightarrow !A_{i_{m}}\,]$ $\mid \hspace {-2pt}\thicksim _{s} !\bot$. Both clauses of Vranas’s step 1 yield $\vdash (C_{i_{1}}\vee ... \vee C_{i_{m}})\leftrightarrow \top$ and $\vdash ((C_{i_{1}}\rightarrow A_{i_{1}})\wedge ... \wedge (C_{i_{m}}\rightarrow A_{i_{m}}))\leftrightarrow \bot$. Then $\vdash (C_{1}\vee ... \vee C_{n})\leftrightarrow \top$ and $\vdash ((C_{1}\rightarrow A_{1})\wedge ... \wedge (C_{n}\rightarrow A_{n}))\leftrightarrow \bot$, so also $\vdash C$ and $\vdash \lnot A$ and $\vdash C\wedge \lnot A$ which was excluded. (The proof for $D\Rightarrow !B$ is done likewise.)

Secondly, due to (1) and (2) all of the following are true:

• (5) ⊢ E → C

• (6) ⊢ E → (A ↔ F)

• (7) ⊢ G → D

• (8) ⊢ G → (B ↔ H)

Since (5) and (6) imply (3), and (7) and (8) imply (4), both (ii) and the left-to-right direction of (iii) are proved as for weak inferences. It remains to prove the right-to-left direction of (iii)

• (iv) $\vdash (E\vee G)\rightarrow (((E\rightarrow F)\wedge (G\rightarrow H))\rightarrow ((C\rightarrow A)\wedge (D\rightarrow B)))$.

For a strong deduction, restriction (R) must hold. Suppose it holds because the conditions of the heads of (DisCo) are logically equivalent:

• (9) ⊢ E ↔ G

Then (iv) becomes equivalent to

$$ \vdash E\rightarrow(((E\rightarrow F)\wedge(E\rightarrow H))\rightarrow ((C\rightarrow A)\wedge (D\rightarrow B))), $$

(4)

which is propositionally equivalent to

$$ \vdash E\rightarrow((F\wedge H)\rightarrow ((C\rightarrow A)\wedge (D\rightarrow B))). $$

(5)

From (6) we have ⊢ E → (A ↔ F), and from (8), (9) ⊢ E → (B ↔ H), so (iv) reduces further to

$$ \vdash E\rightarrow((A\wedge B)\rightarrow ((C\rightarrow A)\wedge (D\rightarrow B))), $$

(6)

which holds trivially.

Suppose restriction (R) holds because above neither head of (DisCo) the rule (SA) is applied. We are examining case B that also excludes any application of the rule (EFQ) above any of the heads. But all other rules that may have been applied above the heads result only in tails whose conditions are logically implied by those of their heads. So the induction hypothesis additionally supplies:

• (10) $~\vdash C\rightarrow E$

• (11) $~\vdash D\rightarrow G$

From (5) and (10) we obtain $\vdash C\leftrightarrow E$, and from (7) and (11) we obtain $\vdash D\leftrightarrow G$. So (iv) is equivalent to

$$ \vdash (C\vee D)\rightarrow(((C\rightarrow F)\wedge(D\rightarrow H))\rightarrow ((C\rightarrow A)\wedge (D\rightarrow B))). $$

(7)

From (6) and (10) we have $\vdash C\rightarrow (A\leftrightarrow F)$, and from (8) and (11) we have $\vdash D\rightarrow (B\leftrightarrow H)$, so (iv) is equivalent to

$$ \vdash (C\vee D)\rightarrow(((C\rightarrow A)\wedge(D\rightarrow B))\rightarrow ((C\rightarrow A)\wedge (D\rightarrow B))), $$

(8)

which holds trivially. □