# Destructive dilemma

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Destructive dilemma

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)