Rule of inference of propositional logic
Destructive dilemmaType | Rule of inference |
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Field | Propositional calculus |
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Statement | If implies and implies and either is false or is false, then either or must be false. |
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Symbolic statement | ![{\displaystyle {\frac {P\to Q,R\to S,\neg Q\lor \neg S}{\therefore \neg P\lor \neg R}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b97149a85956a6f31c6a24308f9b5e6cbc01099e) |
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Transformation rules |
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Propositional calculus |
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Rules of inference |
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- Implication introduction / elimination (modus ponens)
- Biconditional introduction / elimination
- Conjunction introduction / elimination
- Disjunction introduction / elimination
- Disjunctive / hypothetical syllogism
- Constructive / destructive dilemma
- Absorption / modus tollens / modus ponendo tollens
- Negation introduction
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Rules of replacement |
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- Associativity
- Commutativity
- Distributivity
- Double negation
- De Morgan's laws
- Transposition
- Material implication
- Exportation
- Tautology
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Predicate logic |
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Rules of inference |
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- Universal generalization / instantiation
- Existential generalization / instantiation
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Destructive dilemma[1][2] is the name of a valid rule of inference of propositional logic. It is the inference that, if P implies Q and R implies S and either Q is false or S is false, then either P or R must be false. In sum, if two conditionals are true, but one of their consequents is false, then one of their antecedents has to be false. Destructive dilemma is the disjunctive version of modus tollens. The disjunctive version of modus ponens is the constructive dilemma. The destructive dilemma rule can be stated:
![{\displaystyle {\frac {P\to Q,R\to S,\neg Q\lor \neg S}{\therefore \neg P\lor \neg R}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b97149a85956a6f31c6a24308f9b5e6cbc01099e)
where the rule is that wherever instances of "
", "
", and "
" appear on lines of a proof, "
" can be placed on a subsequent line.
The destructive dilemma rule may be written in sequent notation:
![{\displaystyle (P\to Q),(R\to S),(\neg Q\lor \neg S)\vdash (\neg P\lor \neg R)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f37282c24ff5d59f0a8a4c7b08ee63871852796f)
where
is a metalogical symbol meaning that
is a syntactic consequence of
,
, and
in some logical system;
and expressed as a truth-functional tautology or theorem of propositional logic:
![{\displaystyle (((P\to Q)\land (R\to S))\land (\neg Q\lor \neg S))\to (\neg P\lor \neg R)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18c50cfc3ec414f0e2a5f77299ebba48c1bea43d)
where
,
,
and
are propositions expressed in some formal system.
Natural language example
- 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, or both.
- Therefore, either it will not rain, or it will not be sunny, or both.
Proof
Step | Proposition | Derivation |
1 | ![{\displaystyle P\to Q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7cad5b2c2991ae1dbded560c5d875fbf49fe8ea) | Given |
2 | ![{\displaystyle R\to S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/499bc5e6b1153b43ba86d31780bc215f01aa0bf1) | Given |
3 | ![{\displaystyle \neg Q\lor \neg S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fa62b44c2fc97e98de5571a1f1de92a34f6b386) | Given |
4 | ![{\displaystyle \neg Q\to \neg P}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0777042bd6b742c2a329ba3c5223962d9b66c0c6) | Transposition (1) |
5 | ![{\displaystyle \neg S\to \neg R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19aef7d1f6dbdf4a2024bb7132568c4603a21eab) | Transposition (2) |
6 | ![{\displaystyle (\neg Q\to \neg P)\land (\neg S\to \neg R)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d170190931d2550a8f4975cdb86ade44132af7f) | Conjunction introduction (4,5) |
7 | ![{\displaystyle \neg P\lor \neg R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15e09e93ed06ec5e642037d7edfc2e3b235f7e82) | Constructive dilemma (6,3) |
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. | ![{\displaystyle ((P\to Q)\land (R\to S))\land (\neg Q\lor \neg S)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c88fe46c208742bd49f11eb2c03ac761e37696fb) | (CP assumption) |
2. | ![{\displaystyle (P\to Q)\land (R\to S)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44806c1f158ecd7465f142ec447194e96eafc57f) | (1: simplification) |
3. | ![{\displaystyle P\to Q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7cad5b2c2991ae1dbded560c5d875fbf49fe8ea) | (2: simplification) |
4. | ![{\displaystyle R\to S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/499bc5e6b1153b43ba86d31780bc215f01aa0bf1) | (2: simplification) |
5. | ![{\displaystyle \neg Q\lor \neg S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fa62b44c2fc97e98de5571a1f1de92a34f6b386) | (1: simplification) |
6. | ![{\displaystyle \neg (\neg P\lor \neg R)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b664421860f0acfa7439bf7e64587aaf02301e7c) | (RAA assumption) |
7. | ![{\displaystyle \neg \neg P\land \neg \neg R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b16f5e3827d1277a5dc39cf97ce888fd297c0d7d) | (6: De Morgan's Law) |
8. | ![{\displaystyle \neg \neg P}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53d7b8bafb9762c3a07924b6c06be6e08cb5680f) | (7: simplification) |
9. | ![{\displaystyle \neg \neg R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f92dc03db459735964ea9d367f09bc9e8a083b3) | (7: simplification) |
10. | ![{\displaystyle P}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4dc73bf40314945ff376bd363916a738548d40a) | (8: double negation) |
11. | ![{\displaystyle R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b0bfb3769bf24d80e15374dc37b0441e2616e33) | (9: double negation) |
12. | ![{\displaystyle Q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8752c7023b4b3286800fe3238271bbca681219ed) | (3,10: modus ponens) |
13. | ![{\displaystyle S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4611d85173cd3b508e67077d4a1252c9c05abca2) | (4,11: modus ponens) |
14. | ![{\displaystyle \neg \neg Q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac553ce35ffa9e2927b2c80ace86053e9711476f) | (12: double negation) |
15. | ![{\displaystyle \neg S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41fbfa293f2fa92b1251e98c56f92eb45c3f4a08) | (5, 14: disjunctive syllogism) |
16. | ![{\displaystyle S\land \neg S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d39ae52334d0889832d8cf09b1ab83ac618fdee) | (13,15: conjunction) |
17. | ![{\displaystyle \neg P\lor \neg R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15e09e93ed06ec5e642037d7edfc2e3b235f7e82) | (6-16: RAA) |
|
18. | ![{\displaystyle (((P\to Q)\land (R\to S))\land (\neg Q\lor \neg S)))\to \neg P\lor \neg R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/afa96f1d6c284fc75cdb79dd266652139ece406a) | (1-17: CP) |
References
- ^ Hurley, Patrick. A Concise Introduction to Logic With Ilrn Printed Access Card. Wadsworth Pub Co, 2008. Page 361
- ^ Moore and Parker
Bibliography
- Howard-Snyder, Frances; Howard-Snyder, Daniel; Wasserman, Ryan. The Power of Logic (4th ed.). McGraw-Hill, 2009, ISBN 978-0-07-340737-1, p. 414.
External links
- http://mathworld.wolfram.com/DestructiveDilemma.html