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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.

1 Answer 1

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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 
5
  • I just found that P, Q1 & Q2 |- S if and only if P, Q1 |- (Q2 -> S). This is called exportation.
    – Tim
    Oct 2, 2023 at 12:50
  • Thanks. When is P, Q1 & Q2 |- S if and only if P, Q1 |- S or P, Q2 |- S? See the first case in philosophy.stackexchange.com/questions/103482/…
    – Tim
    Oct 2, 2023 at 12:50
  • You can always go from P, Q1 ⊢ S to P, Q1, Q2 ⊢ S, or from P, Q2 ⊢ S to P, Q1, Q2 ⊢ S. That follows from the monotonicity of entailment, which is a property of classical logic. You cannot go in the converse direction, unless you also have Q1 ∨ Q2 ⊢ Q1 ∧ Q2.
    – Bumble
    Oct 2, 2023 at 21:41
  • Thanks. What does "Q1 ∨ Q2 ⊢ Q1 ∧ Q2" mean? Does it hold in the first case in philosophy.stackexchange.com/questions/103482/…?
    – Tim
    Oct 2, 2023 at 22:35
  • It means that from the disjunction of Q1, Q2, we can prove the conjunction of Q1, Q2. Whether this is so depends on your Q1 and Q2. It might be better just to use an object language conditional, (Q1 ∨ Q2) → (Q1 ∧ Q2). If that sentence holds true, then you can make the converse inference. It is not a given in your example.
    – Bumble
    Oct 2, 2023 at 22:44

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