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1. 2 Day 14 Day 13 Day 12 Day 11 Day 10 Day 09 Introductory Material EXAM #2 show: conjunction Indirect Derivation show: atomic show: disjunction Conditional.

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Presentation on theme: "1. 2 Day 14 Day 13 Day 12 Day 11 Day 10 Day 09 Introductory Material EXAM #2 show: conjunction Indirect Derivation show: atomic show: disjunction Conditional."— Presentation transcript:

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2 2 Day 14 Day 13 Day 12 Day 11 Day 10 Day 09 Introductory Material EXAM #2 show: conjunction Indirect Derivation show: atomic show: disjunction Conditional Derivation (CD) Negation Derivation (  D) Direct Derivation (DD)

3 3  6 argument forms, 15 points each, plus 10 free points  Symbolic argument forms (no translations)  For each one, you will be asked to construct a derivation of the conclusion from the premises. rule sheet  The rule sheet will be provided. 1 problemfromSet D 2 problem fromSet E 2 problemsfromSet F 1 problemfromSet G (91-96)

4 4  ––––––  ––––––  DN     –––––––      –––––––  O O  ––––––     ––––––    I I  ––––––      ––––––  O O   ––––––  &    ––––––  &  &I &I –––––––   &  –––––––  &O &O

5 5  D   :  ID   As   :    CD   :    CD   As  :      DD  :  DD          

6 6 Assumption Rule (CD)   If one has a line of the form  :    then one is entitled to write down the formula  on the very next line, as an assumption. Assumption Rule (  D)   If one has a line of the form  :   then one is entitled to write down the formula  on the very next line, as an assumption. Contradiction-In (  I) if you have a formula   and you have its negation  then you are entitled to infer––––  a contradiction (absurdity) 

7 7  :  ° ° °  DD In Direct Derivation (DD), one directly arrives at the very formula one is trying to show.

8 8  :     As  :  ° ° ° CD ??

9 9  :    As   :  ° °  DD DD

10 10 (1)P  QPr (2)  P  QPr (3)  : Q?? We are stuck!!  we haveP  Q  O so to apply  O we must findP or find  Q  we also have  P  Q  O so to apply  O we must find  P or find  Q 

11 11  :    As   :  ° °  IDID DD  D This is exactly parallel to  D, and is another version of the traditional mode of reasoning known as REDUCTIO AD ABSURDUM  :    As   :  ° °  DD DD

12 12 any Although ID can, in principle, be used on any formula, it is best used on two types of formulas. 1.atomic formulasP, Q, R, etc. 2.disjunctions    ID  D The difference between ID and  D is that  Dnegations  D applies only to negations, IDall formulas whereas ID applies (in principle) to all formulas; it is a generic rule, like direct-derivation.

13 13  :    As   :  ° °  IDID DD  is atomic (P,Q,R, etc.)

14 14 (8) (7) (6) (5) (4) (3) (2) (1) 4,7,  2,6, Q 1,4,  P DD  :  As  Q ID  : Q Pr  P  Q Pr P  Q II OO OO

15 15 (8) (7) (6) (5) (4) (3) (2) (1) 1,7,  3,5, P &  Q DD  :  As  Q ID  : Q As P CD  : P  Q Pr  (P &  Q) II &I&I

16 16  :     [    ]As   :  ° °  IDID DD

17 17          ––––––––––––––––––    

18 18 (8) (7) (6) (5) (4) (3) (2) (1) 6,7,  1,5, Q  Q 3,  P DD  :  As  (P  Q) ID  : P  Q Pr  P  Q OO II OO

19 19 (11) (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) 3,10,  8,9,  Q 1,7,  Q  R  R 5,  P DD  :  As  (P  R) ID  : P  R As Q CD  : Q  (P  R) Pr  P  (  Q  R) OO II OO OO

20 20 (9) (8) (6) (5) 7, (11) (10) (7) (4) (3) (2) (1) 6,10,  8,9,  P &  Q  Q  P 1,5,  (P  Q)  (  P &  Q) OO  (P & Q) DD  :  As  [ (P & Q)  (  P &  Q)] ID  : (P & Q)  (  P &  Q) Pr (P  Q)  (P & Q) 3, OO II &I&I OO

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