2. 1

3. 2 We demonstrate (show) that an argument is valid deriving by deriving (deducing) its conclusion from its premises using a few fundamental modes of reasoning.

4. 3    Modus (Tollendo) Tollens (denying by denying)      ––––––    Modus (Ponendo) Ponens (affirming by affirming)     ––––––    Modus Tollendo Ponens (1) (affirming by denying)      ––––––    Modus Tollendo Ponens (2) (affirming by denying)      ––––––  aka disjunctive syllogism

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

6. 5 Tek gives Manny a high-fiveTek gives A-Rod a high-five

7. 6 available on course web page (textbook) provided on exams keep this in front of you when doing homework don’t make up your own rules

8. 7

9. 8 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

10. 9 conjunction  &  if you have a conjunction  &  then you are entitled to infer–––––– first conjunct  its first conjunct  conjunction  &  if you have a conjunction  &  then you are entitled to infer–––––– second conjunct  its second conjunct  ‘have’ means ‘have as a whole line’ rules apply only to whole lines not pieces of lines

11. 10 formula  if you have a formula  formula  and you have a formula  then you are entitled to infer–––––– conjunction  &  their (first) conjunction  &  formula  if you have a formula  formula  and you have a formula  conjunction  &  then you are entitled to infer–––––– their (second) conjunction  & 

12. 11 disjunction    if you have a disjunction    negation1 st disjunct   and you have the negation of its 1 st disjunct   then you are entitled to infer –––––– second disjunct  its second disjunct  disjunction    if you have a disjunction    negation2 nd disjunct   and you have the negation of its 2 nd disjunct   then you are entitled to infer –––––– first disjunct  its first disjunct  what we earlier called ‘modus tollendo ponens’

13. 12 formula  if you have a formula  then you are entitled to infer–––––– disjunctionany    its disjunction with any formula to its right    formula  if you have a formula  then you are entitled to infer–––––– disjunctionany    its disjunction with any formula to its left   

14. 13 conditional    if you have a conditional    antecedent  and you have its antecedent  then you are entitled to infer –––––– consequent  its consequent  conditional    if you have a conditional    negationconsequent   and you have the negation of its consequent   then you are entitled to infer –––––– negationantecedent   the negation of its antecedent   What we earlier called ‘modus ponens’ and ‘modus tollens’

15. 14 THERE IS NO SUCH THING AS ARROW-IN (  I)  what we have instead is CONDITIONAL DERIVATION (CD) which we examine later

16. 15 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 lines not pieces of lines

17. 16  :  ° ° °  In Direct Derivation (DD), one directly arrives at DD The Original and Fundamental  -Rule the very formula one is trying to show.

18. 17 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) 5,9,  Q 4,8,  P 3,7,  T 1,2,  R DD  :  Q Pr  P   Q Pr P  T Pr  R   T Pr R  S Pr SS  S ; R  S ;  R   T ; P  T ;  P   Q /  Q OO OO OO OO

19. 18  If you have a line of the form   , arrow-out  O then try to apply arrow-out (  O), which requires a second formula as input, in particular, either  or .      deduce find have or

20. 19 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) 5,9,  Q 4,8,  P 3,7,  T 1,2,  R DD  :  Q Pr  P   Q Pr  P  T Pr R   T Pr R  S Pr SS  S ; R  S ; R   T ;  P  T ;  P   Q /  Q OO OO OO OO

21. 20  If you have a line of the form   , wedge-out  O then try to apply wedge-out (  O), which requires a second formula as input, in particular, either   or .      deduce find have or

22. 21 (8) (7) (6) (5) (4) (3) (2) (1) 5,7, R 3,6, Q  R 1,5, P  Q 1,2,  R DD  :  Q Pr (P  Q)  (Q  R) Pr [(P  Q)  R]   R Pr (P  Q)  R (P  Q)  R ; [(P  Q)  R]   R ; (P  Q)  (Q  R) /  Q OO OO OO OO

23. 22 (8) (7) (6) (5) (4) (3) (2) (1) 2,7, T 6, R  S 1,5, S 3, P  Q DD  : T PrP (R  S)  T Pr (P  Q)  S (P  Q)  S ; (R  S)  T) ; P / T OO II OO II

24. 23 If you need  then look for either disjunct: find  OR find  then v I apply v I  to get   

25. 24 (8) (9) (10) (11) (12) (7) (6) (5) (4) (3) (2) (1) 1,7, Q 3,6,  R 2,9, Q  S 8,10, S 8,11, Q & S P 4,&O  T DD  Q & S Pr  T & P Pr R  T Pr R  (Q  S) Pr P  Q P  Q ; R  (Q  S) ; R  T ;  T & P / Q & S OO OO OO OO &I&I

26. 25 If you have a line of the form &&,&&, ampersand-out then apply ampersand-out (&O), which can be applied immediately to produce and 

27. 26 If you are trying to find or show &&&& then look for both conjuncts: find  AND find  then & I apply & I & to get  & 

28. 27