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INTRO LOGIC

INTRO LOGIC. Derivations in SL 2. DAY 10. Review. We demonstrate (show) that an argument is valid by deriving ( deducing ) its conclusion from its premises using a few fundamental modes of reasoning . Initial Modes of Reasoning.

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INTRO LOGIC

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  1. INTRO LOGIC Derivations in SL2 DAY 10

  2. Review We demonstrate (show) that an argument is valid by deriving (deducing) its conclusion from its premises using a few fundamental modes of reasoning.

  3. Initial Modes of Reasoning Modus (Ponendo) Ponens(affirming by affirming) –––––– Modus (Tollendo) Tollens(denying by denying)–––––– Modus Tollendo Ponens (1)(affirming by denying)–––––– Modus Tollendo Ponens (2)(affirming by denying)–––––– aka disjunctive syllogism

  4. Example 1 (review) S ; RS ; RT ; PT ; PQ / Q 1. write all premises (1) S Pr (2) R  S Pr (3) R T Pr 2. write ‘show’ conclusion (4) P  T Pr (5) P Q Pr 3. apply rules to available lines (6) : Q DD (7) R 1,2, MT (8) T 3,7, MTP1 4. box and cancel(high-five!) (9) P 4,8, MTP2 (10) Q 5,9, MP

  5. other high-fives Tek gives Manny a high-five Tek gives A-Rod a high-five

  6. Rule Sheet provided on exams don’t makeupyourownrules available on courseweb page(textbook) keep this in front of you when doing homework

  7. Rules discussed Today

  8. Rules of Inference – Basic Idea • Every connective has • (1) a take-out rule • also called an: • elimination-rule • OUT-rule • (2) a put-in rule • also called an: • introduction-rule • IN-rule

  9. if you have a conjunction & then you are entitled to infer –––––– its first conjunct  Ampersand-Out (&O) if you have a conjunction & then you are entitled to infer –––––– its second conjunct  ‘have’ means ‘have as a whole line’ rules apply only to whole linesnot pieces of lines

  10. if you have a formula  and you have a formula  then you are entitled to infer –––––– their (first) conjunction& Ampersand-IN (&I) if you have a formula  and you have a formula  then you are entitled to infer ––––––their (second) conjunction &

  11. if you have a disjunction and you have the negation of its 1st disjunct  then you are entitled to infer –––––– its second disjunct  Wedge-Out (O) if you have a disjunction and you have the negation of its 2nd disjunct  then you are entitled to infer –––––– its first disjunct  what we earlier called ‘modus tollendo ponens’

  12. if you have a formula  then you are entitled to infer –––––– its disjunction with any formula to its right  Wedge-IN (I) if you have a formula  then you are entitled to infer –––––– its disjunction with any formula to its left  

  13. if you have a conditional and you have its antecedent  then you are entitled to infer –––––– its consequent  Arrow-Out (O) if you have a conditional and you have the negation of its consequent  then you are entitled to infer –––––– the negation of its antecedent  What we earlier called ‘modus ponens’ and ‘modus tollens’

  14. THERE IS NO SUCH THING ASARROW-IN (I)  Arrow-Introduction what we have instead is CONDITIONAL DERIVATION (CD)  which we examine later

  15. Double-Negation (DN) • if you have a formula  • then you are entitled to infer ––––– • its double-negative  if you have a double-negative  then you are entitled to infer ––––– the formula  rules apply only to whole linesnot pieces of lines

  16. Direct Derivation The Original and Fundamental -Rule • :  • ° • ° • ° •  DD In Direct Derivation (DD),one directly arrives at the very formula one is trying to show.

  17. Example 2 S;R  S;R T;P  T;P Q/Q (1) S Pr (2) R  S Pr (3) R T Pr (4) P  T Pr (5) P Q Pr (6) : Q DD (7) R 1,2, O (8) T 3,7, O (9) P 4,8, O (10) Q 5,9, O

  18. Arrow-Out Strategy • If you have a line of the form , • then try to apply arrow-out (O), • which requires a second formula as input, • in particular, either  or . have    find or deduce  

  19. Example 3 S;R  S;R T;P  T;P Q/Q (1) S Pr (2) R  S Pr (3) R T Pr (4) P  T Pr (5) P Q Pr (6) : Q DD (7) R 1,2, O (8) T 3,7, O (9) P 4,8, O (10) Q 5,9, O

  20. Wedge-Out Strategy • If you have a line of the form , • then try to apply wedge-out (O), • which requires a second formula as input, • in particular, either  or . have    find or deduce  

  21. Example 4 (P  Q)  R;[(P  Q)  R] R;(P  Q)  (Q  R)/Q (1) (P  Q)  R Pr (2) [(P  Q)  R] R Pr (3) (P  Q)  (Q  R) Pr (4) : Q DD (5) R 1,2, O (6) P  Q 1,5, O (7) Q  R 3,6, O (8) R 5,7, O

  22. Example 5 (P  Q) S ; (R  S) T) ; P /T (1) (P  Q) S Pr (2) (RS) T Pr (3) P Pr (4) : T DD (5) P  Q 3, I (6) S 1,5, O (7) R  S 6, I (8) T 2,7, O

  23. Wedge-In Strategy • If you need •  • then look for either disjunct: • find  • OR • find  • then • apply vI • to get 

  24. Example 6 P  Q ; R  (Q  S) ; R  T ;T & P / Q & S (1) P  Q Pr (2) R  (Q  S) Pr (3) R  T Pr (4) T & P Pr (5) Q & S DD (6) T 4,&O (7) P (8) Q 1,7, O (9) R 3,6, O (10) Q  S 2,9, O (11) S 8,10, O (12) Q & S 8,11, &I

  25. Ampersand-Out Strategy • If you have a line of the form • &, • then apply ampersand-out (&O), • which can be applied immediatelyto produce •  • and • 

  26. Ampersand-In Strategy • If you are trying to find or show • & • then look for both conjuncts: • find  • AND • find  • then • apply &I • to get &

  27. THE END

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