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2. 2 Exam 1Sentential LogicTranslations (+) Exam 2Sentential LogicDerivations Exam 3Predicate LogicTranslations Exam 4Predicate LogicDerivations 6 derivations@ 15 points+ 10 free points Exam 5very similar to Exam 3 Exam 6very similar to Exam 4 + + +

3. 3 When computing your final grade, four highest scores I count your four highest scores. missedzero (A missed exam counts as a zero.) When computing your final grade, four highest scores I count your four highest scores. missedzero (A missed exam counts as a zero.)

4. 4 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

5. 5  DD  ID  CD   D  &D  etc.  DD  ID  CD DD  &D  etc.  &I  &O  vO   O    O  etc.  &I  &O  vO OO OO  etc.

6. 6 OO Tilde-Universal-Out OO Universal-Out UDUniversal Derivation OO Tilde-Existential-Out OO Existential-Out II Existential-In day 3 day 1 day 2 day 3 day 2 day 1

7. 7       –––––         –––––        is an OLD name (more about this later) OO II

8. 8 UDUniversal Derivation OO Existential-Out

9. 9 … (?) (c) (?) (b) (?) (a) (3) (2) (1) every F is H ; everyone is F / everyone is H … … ??  : Hc ??  : Hb ??  : Ha ??  :  xHx Pr  xFx Pr  x(Fx  Hx) (3)&.&.&.D  : Ha & Hb & Hc & ……… what is ultimately involved in showing a universal

10. 10 (7) (6) (5) (4) (3) (2) (1) 5,6, Ha 2, Fa 1, Fa  Ha DD  : Ha ??  :  xHx Pr  xFx Pr  x(Fx  Hx) one down, a zillion to go! OO OO OO

11. 11 (7) (6) (5) (4) (3) (2) (1) 5,6, Hb 2, Fb 1, Fb  Hb DD  : Hb ??  :  xHx Pr  xFx Pr  x(Fx  Hx) two down, a zillion to go! OO OO OO

12. 12 (7) (6) (5) (4) (3) (2) (1) 5,6, Hc 2, Fc 1, Fc  Hc DD  : Hc ??  :  xHx Pr  xFx Pr  x(Fx  Hx) three down, a zillion to go! OO OO OO

13. 13 DD  : Hc (4) 1,  O Fc  Hc (5) 2,  O Fc(6) 5,6,  O Hc(7) DD  : Hb (4) 1,  O Fb  Hb (5) 2,  O Fb(6) 5,6,  O Hb(7) 5,6,  O Ha(7) 2,  O Fa(6) 1,  O Fa  Ha (5) DD  : Ha (4)

14. 14 So, all we need to do is do one derivation with one name (say, ‘a’) and then argue that all the other derivations will look the same. To ensure this, we must ensure that the name is general, which we can do by making sure the name we select is NEW. a name counts as NEW precisely if it occurs nowhere in the derivation unboxed or uncancelled

15. 15  :        :    ° ° ° ° UD ?? i.e., one that is occurs nowhere in the derivation unboxed or uncancelled must be a NEW name, replaces      is any (official) formula  is any variable

16. 16 OLD name       –––––     NEW name  :        :    a name counts as OLD precisely if it occurs somewhere in the derivation unboxed and uncancelled a name counts as NEW precisely if it occurs nowhere in the derivation unboxed or uncancelled OO UD

17. 17 (8) (7) (6) Ha Fa Fa  Ha 6,7, 3, 1, (5) (4) (3) (2) (1)  : Ha  :  xHx  xFx  :  xFx   xHx  x(Fx  Hx) DD UD As CD Pr every F is H / if everyone is F, then everyone is H a new a old OO OO OO

18. 18 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) every F is G ; every G is H / every F is H Ha Ga Ga  Ha Fa  Ga  : Ha Fa  : Fa  Ha  :  x(Fx  Hx)  x(Gx  Hx)  x(Fx  Gx) 8,9, 5,7, 2, As CD UD Pr a new a old 1, DD OO OO OO OO

19. 19 (6) (5) (4) (3) (2) (1) everyone R’s everyone / everyone is R’ed by everyone 5, Rba 1,  yRby DD  : Rba UD  :  yRya UD  :  x  yRyx Pr  x  yRxy b new a new b old a old OO OO

20. 20 any variable (z, y, x, w …) any NEW name (a, b, c, d, …) any formula replaces        –––––   

21. 21 OLD name       ––––––     NEW name       ––––––    a name counts as OLD precisely if it occurs somewhere unboxed and uncancelled a name counts as NEW precisely if it occurs nowhere unboxed or uncancelled OO OO

22. 22 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) every F is un-H / no F is H   Ha Ha Fa Fa   Ha Fa & Ha DD  :  As  x(Fx & Hx) DD  :   x(Fx & Hx) Pr  x(Fx   Hx) new old 8,9, 6,7, 5, 1, 3, II OO &O OO OO

23. 23 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) some F is not H / not every F is H  Ha  Ha Fa Fa  Ha Fa &  Ha DD  :  As  x(Fx  Hx) ID  :   x(Fx  Hx) Pr  x(Fx &  Hx) 8,9, 6,7, 5, 3, 1, II OO &O OO OO new old

24. 24 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) every F is G ; some F is H / some G is H 9,  x(Gx & Hx) 7,8, Ga & Ha 5,6, Ga Ha 4, Fa 1, Fa  Ga 2, Fa & Ha DD  :  x(Gx & Hx) Pr  x(Fx & Hx) Pr  x(Fx  Gx) II &I&I OO &O OO OO new old

25. 25 (9) (8) (7) (6) (5) (4) (3) (2) (1) if anyone is F then everyone is H / if someone is F, then everyone is H 8, Ha 6,7,  xHx 1, Fb   xHx 3, Fb DD  : Ha UD  :  xHx As  xFx CD  :  xFx   xHx Pr  x(Fx   xHx) OO OO OO OO new old new old

26. 26 (9) (8) (7) (6) (5) (4) (3) (2) (1) if someone is F, then everyone is H / if anyone is F then everyone is H 8, Hb 1,7,  xHx 4,  xFx DD  : Hb UD  :  yHy As Fa CD  : Fa   yHy UD  :  x(Fx   yHy) Pr  xFx   xHx OO OO II new old new

27. 27 (7) (6) (5) (4) (3) (2) (1) someone R’s someone ??missing premises?? / everyone R’s everyone 6, Rcd 1,  yRcy ??  : Rab UD  :  yRay UD  :  x  yRxy Pr??? Pr  x  yRxy (8) ?? OO OO new

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