What is the justification of a complex dilemma?

Question

7.8 The Dilemma in Copi's Introduction to Logic says:

Complex dilemma: An argument consisting of (a) a disjunction, (b) two conditional premises linked by a conjunction, and (c) a conclusion that is not a single categorical proposition (as in a simple dilemma) but a disjunction, a pair of (usually undesirable) alternatives.

Does the above mean that a complex dilemma has the following form:

P | Q, (P->R) & (Q->S)  |-  R | S

?

When trying to justify the form, I got some contradictory results in two ways.

Before moving to the justification, let me introduce the following two inference rules. Are they both correct?

1. To prove P, Q1 & Q2 |- S, i.e. with conjunctive in antecedent, we can either

   P, Q1 |- S
   --------------------------
   P, Q1 & Q2 |- S
   

   or

   P, Q2 |- S
   --------------------------
   P, Q1 & Q2 |- S
   

2. To prove P, Q1 | Q2 |- S, i.e. with disjunctive in antecedent, we can

   P, Q1 |- S,  
   P, Q2 |- S
   --------------------------
   P, Q1 | Q2 |- S
   

Now to prove P | Q, (P->R) & (Q->S) |- R | S, the form of a complex dilemma, let me try to approach it in two ways.

1. Consider the first way. According to the first inference rule above, it suffices to show that either

   P | Q, P->R  |-  R | S
   

   or

   P | Q, Q->S  |-  R | S
   

   To show P | Q, P->R |- R | S, according to the second inference rule above, it suffices to show

   P, P->R  |-  R | S
   

   and

   Q, P->R  |-  R | S
   

   the second of which doesn't hold, neither does P | Q, P->R |- R | S.

   Similarly, we can't show P | Q, Q->S |- R | S. Thus, we can't show P | Q, (P->R) & (Q->S) |- R | S in this way.

2. Consider the second way to prove P | Q, (P->R) & (Q->S) |- R | S. According to the second inference rule above, it suffices to show that either

   P, (P->R) & (Q->S)  |-  R | S
   

   and

   Q, (P->R) & (Q->S)  |-  R | S
   

   To show P, (P->R) & (Q->S) |- R | S, according to the first inference rule above, it suffices to show

   P, P->R  |-  R | S
   

   or

   P, Q->S  |-  R | S
   

   the first of which holds, so does P, (P->R) & (Q->S) |- R | S.

   Similarly, we can show Q, (P->R) & (Q->S) |- R | S. Thus, P | Q, (P->R) & (Q->S) |- R | S holds.

Thanks.

Answer 1

You have the correct representation of the form of the argument. It is also sometimes called constructive dilemma.

Your first inference rule states a sufficient condition for proving S, but not a necessary one. In order to prove P, Q1 & Q2 ⊢ S we may need both Q1 and Q2. If we can prove S from either one alone, well and good, but in general we may need both together. This is why your attempt to derive a proof under 1 fails.

When asking for a justification of a rule in logic, it makes sense to ask what you would count as a proof. You could just construct a truth table. You could write a proof using some natural deduction rules, but you might object that these are just as much in need of justification. Here is a proof, writing ∨ for disjunction, ∧ for conjunction, and → for the material conditional.

1. P ∨ Q                Premise
2. P → R                Premise
3. Q → S                Premise 
4.  | ¬(R ∨ S)          Assumption, for the purpose of reductio
5.  | ¬R ∧ ¬S           From 4, by de Morgan's rules
6.  | ¬R                From 5, ∧-elimination
7.  | ¬P                From 2, 6, modus tollens
8.  | ¬S                From 5, ∧-elimination
9.  | ¬Q                From 3, 8, modus tollens
10. | Q                 From 1, 7, disjunctive syllogism
11. | Q ∧ ¬Q            From 9, 10, ∧-introduction
12. ¬¬(R ∨ S)           From 4, 11, by reductio 
13. R ∨ S               From 12, double negation elimination