ln2-page11 - 5 ⇠_ O 5 ⇠_ O 8 DN 1,9 ! O 2,7 ! O 10 DN...

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For “and” and “iff” statements, break the problem into two $ I. SHOW line Form Try Assume SHOW conjunction p q DD 1st: SHOW: p ID p 6 then: SHOW: q ID q 6 biconditional p $ q DD 1st: SHOW: p ! q CD p q then: SHOW: q ! p CD q p This should tell you how to set up any problem: Look at your SHOW line, ±gure out which kind of derivation works best and then make your assumption and write your new SHOW line. Look at your new SHOW line and repeat the procedure until you get SHOW 6 . Now, use the DD rules to get a contradiction. This will work for almost all problems. Examples: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. ( P _ Q ) ! P ( R P ) !⇠ R SHOW: P R SHOW: P P SHOW: 6 P _ Q P 6 SHOW: R R SHOW: 6 R P R 6 P R Pr Pr DD ID Ass DD 5 _ I 1,7 ! O 5,8 6 I ID Ass DD 2,13 ! O 11,14 6 I [ Remember you can use a SHOW line after SHOW has been crossed out. ] 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. A ! B C ! D B _⇠ D SHOW: C _⇠ A ( C _⇠ A ) SHOW: 6 C ⇠⇠ A A B D ⇠⇠ B D 6 Pr Pr Pr ID Ass DD
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Unformatted text preview: 5 ⇠_ O 5 ⇠_ O 8 DN 1,9 ! O 2,7 ! O 10 DN 3,12 _ O 11,13 6 I 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. P $ ( ⇠ R ! R ) ⇠ ⇠ ⇠ SHOW: ( P _⇠ Q ) $ ( Q ! R ) ⇠ ⇠ ⇠ SHOW: ( P _⇠ Q ) ! ( Q ! R ) P _⇠ Q ⇠ ⇠ ⇠ SHOW: Q ! R Q ⇠ ⇠ ⇠ SHOW: R ⇠ R ⇠ ⇠ ⇠ SHOW: 6 ⇠⇠ Q P P ! ( ⇠ R ! R ) ⇠ R ! R R 6 ⇠ ⇠ ⇠ SHOW: ( Q ! R ) ! ( P _⇠ Q ) Q ! R ⇠ ⇠ ⇠ SHOW: P _⇠ Q ⇠ ( P _⇠ Q ) ⇠ ⇠ ⇠ SHOW: 6 ⇠ P ⇠⇠ Q ( ⇠ R ! R ) ! P ⇠ ( ⇠ R ! R ) ⇠ R & ⇠ R ⇠ R ⇠ Q 6 ( P _⇠ Q ) $ ( Q ! R ) Pr DD CD Ass CD Ass ID Ass DD 6 DN 4,10 _ O 1 $ O 11,12 ! O 8,13 ! O 8,14 6 I CD Ass ID Ass DD 19 ⇠_ O 19 ⇠_ O 1 $ O 21,23 ! O 24 ⇠! O 25 &O 17,26 ! O 22,27 6 I 3,16 $ I 11...
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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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