Destructive dilemma

Destructive dilemma
Rules of inference
Propositional calculus
Modus ponens (A→B, A ⊢ B)
Modus tollens (A→B, ¬B ⊢ ¬A)
Modus ponendo tollens (¬(A∧B), A ⊢ ¬B)
Conjunction introduction (A, B ⊢ A∧B)
Simplification (A∧B ⊢ A)
Disjunction introduction (A ⊢ A∨B)
Disjunction elimination (A∨B, A→C, B→C ⊢ C)
Disjunctive syllogism (A∨B, ¬A ⊢ B)
Hypothetical syllogism (A→B, B→C ⊢ A→C)
Constructive dilemma (A→P, B→Q, A∨B ⊢ P∨Q)
Destructive dilemma (A→P, B→Q, ¬P∨¬Q ⊢ ¬A∨¬B)
Biconditional introduction (A→B, B→A ⊢ A↔B)
Biconditional elimination (A↔B ⊢ A→B)
Predicate calculus
Universal generalization
Universal instantiation
Existential generalization
Existential instantiation
v · d · e

In logic, a destructive dilemma is any logical argument of the following form:

P \rightarrow Q
R \rightarrow S
\neg Q \lor \neg S
\vdash \neg P \lor \neg R

where \vdash represents the logical assertion.

The argument can be read in this way:

  1. If P, then Q
  2. If R, then S
  3. Not Q or not S
  4. Therefore, not P or not R

And to once again restate the argument, one can turn this argument into a conditional, where if the first three premises, then not P or R:

(((P \rightarrow Q) \And (R \rightarrow S)) \And (\neg Q \vee \neg S)) \rightarrow (\neg P \vee \neg R)

The destructive dilemma is the disjunctive version of modus tollens. The disjunctive version of modus ponens is the constructive dilemma. Here is an example of the destructive dilemma in English:

If it rains, we will stay inside.
If it is sunny, we will go for a walk.
Either we will not stay inside, or we will not go for a walk.
Therefore, either it will not rain, or it will not be sunny.

Example proof

The validity of this argument structure can be shown by using both conditional proof (CP) and reductio ad absurdum (RAA) in the following way:

1. ((P \rightarrow Q) \And (R \rightarrow S)) \And (\neg Q \vee \neg S) (CP assumption)
2. (P \rightarrow Q) \And (R \rightarrow S) (1: simplification)
3. (P \rightarrow Q) (2: simplification)
4. (R \rightarrow S) (2: simplification)
5. (\neg Q \vee \neg S) (1: simplification)
6. \neg (\neg P \vee \neg R) (RAA assumption)
7. \neg \neg P \And \neg \neg R (6: DeMorgan's Law)
8. \neg \neg P (7: simplification)
9. \neg \neg R (7: simplification)
10. P (8: double negation)
11. R (9: double negation)
12. Q (3,10: modus ponens)
13. S (4,11: modus ponens)
14. \neg \neg Q (12: double negation)
15. \neg S (5, 14: disjunctive syllogism)
16. S \And \neg S (13,15: conjunction)
17. \neg P \vee \neg R (6-16: RAA)
18. (((P \rightarrow Q) \And (R \rightarrow S)) \And (\neg Q \vee \neg S))) \rightarrow \neg P \vee \neg R (1-17: CP)

References

  • Howard-Snyder, Frances; Howard-Snyder, Daniel; Wasserman, Ryan. The Power of Logic (4th ed.). McGraw-Hill, 2009, p. 414.

External links


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