ln2-page10 - DD? Would you need to do some combination of...

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1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. Q ! P Q ! R SHOW: P _ R ( P _ R ) SHOW: 6 SHOW: Q Q SHOW: 6 P P _ R 6 R P _ R 6 Pr Pr ID Ass DD ID Ass DD 1,7 ! O 9 _ I 4,10 6 I 2,6 ! O 12 _ I 4,13 6 I To make our lives easier, we introduce some special rules dealing with negations of complex statements. ( p _ q ) p ( p _ q ) q ⇠_ O ( p ! q ) p q ⇠! O ( p q ) p !⇠ q ( p $ q ) p $ q With these rules, the above proof simpliFes to: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Q ! P Q ! R SHOW: P _ R ( P _ R ) SHOW: 6 P R Q R 6 Pr Pr ID Ass DD 4 ⇠_ O 4 ⇠_ O 1,6 ! O 2,8 ! O 7,9 6 I Some other derivations that use these rules. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. ( A ! B ) ! C SHOW: B ! C B SHOW: C C SHOW: 6 ( A ! B ) A B B 6 Pr CD Ass ID Ass DD 1,5 ! O 7 ⇠! O 3,9 6 I 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. ( P $ Q ) ! M M _ P ( M Q ) SHOW: M M SHOW: 6 ( P $ Q ) P $ Q P Q !⇠ P ⇠⇠ P Q M !⇠⇠ Q ⇠⇠ Q 6 Pr Pr Pr ID Ass DD 1,5 ! O 7 ⇠$ O 2,5 _ O 8 $ O 9 DN 10,11 ! O 3 5,13 ! O 12,14 6 I H. Strategies and Rules of Thumb When you begin a derivation, you might be unsure of the best way to go about it. Should you do a CD, ID or just try it by
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Unformatted text preview: DD? Would you need to do some combination of these? When should you do one technique within another? The simplest rule is: when in doubt, try an indirect derivation. This is because all problems can be done with ID. Some problems are shorter if you dont use ID, but if you cant see that far ahead, it almost never hurts to try. To make things more speciFc, when it doubt, I recommend constructing a derivation using the following strategies, depending on what kind of statement in your SHOW line: SHOW line orm Try Assume New SHOW atomic statement A ID A 6 negation p ID p 6 disjunction p _ q ID ( p _ q ) 6 conditional p ! q CD p q 10...
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This note was uploaded on 11/22/2011 for the course PHIL 110 taught by Professor Bohn during the Spring '08 term at UMass (Amherst).

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