8.5 --Intro to RAA Proofs Practice with Proofs. Today s Lecture 4/20/10
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1 8.5 --Intro to RAA Proofs Practice with Proofs Today s Lecture 4/20/10
2 Announcements -- Final Exam on May 11 th (now s the time to start studying)! -- Next Tues is the deadline to turn in any late homework -- Homework: - Ex 9.3 pg. 462 Part D (10-20 Even) - Ex 9.3 pgs Part E (5-15 All) - Read
3 Announcements CSUN Philosophy Student Conference Fri., April 23, 2010 Whitsett Room, 4th Floor, Sierra Hall (CSUN campus) 8:30-2:30pm (followed by a barbecue at Northridge Park) Program (available on Tu., April 20th)
4 More on the Final -- Per the syllabus it s cumulative and 40% of grade. Chapter 1 --Be ready to define: valid, invalid, sound argument. --Be ready to recognize whether arguments are sound, valid or invalid.
5 More on the Final -- Per the syllabus it s cumulative and 40% of grade. Chapter 1 --Be ready to define: valid, invalid, sound argument. --Be ready to recognize whether arguments are sound, valid or invalid. Chapter 7: --Be ready to define: counter-example, necessary condition, and sufficient condition. -- Symbolizations (study HW problems; memorize stylistic variants)
6 More on the Final Chapter 8 --Direct proofs (study HW) -- Conditional proofs (study HW)
7 More on the Final Chapter 8 --Direct proofs (study HW) -- Conditional proofs (study HW) Chapter 9: --Symbolizations (study HW) -- Direct proofs (study HW) -- RAA and Conditional Proofs (Study HW)
8 Intro to Reductio ad Absurdum (RAA) Proofs RAA proofs, like conditional proofs, are a type of indirect proof.
9 Intro to Reductio ad Absurdum (RAA) Proofs RAA proofs, like conditional proofs, are a type of indirect proof. Constructing an RAA proof can make it easier to show an argument to be valid.
10 Intro to Reductio ad Absurdum (RAA) Proofs RAA proofs, like conditional proofs, are a type of indirect proof. Constructing an RAA proof can make it easier to show an argument to be valid. RAA proofs are based on the principle that: Any statement that entails a contradiction (i.e. an absurdity) is itself false.
11 Intro to Reductio ad Absurdum (RAA) Proofs RAA proofs, like conditional proofs, are a type of indirect proof. Constructing an RAA proof can make it easier to show an argument to be valid. RAA proofs are based on the principle that: Any statement that implies a contradiction (i.e. an absurdity) is itself false. Why think this principle is true? Well
12 Intro to Reductio ad Absurdum (RAA) Proofs According to the valid argument form Modus Tollens (p! q, ~q " ~p), if a statement entails the truth of another distinct statement, and the distinct statement for whatever reason is false, then it must be the case that that the original statement is false.
13 Intro to Reductio ad Absurdum (RAA) Proofs Now let s say that a statement entails that a contradiction (i.e. a conjunction of a statement and its negation) is true. We could put this symbolically as:
14 Intro to Reductio ad Absurdum (RAA) Proofs Now let s say that a statement entails that a contradiction (i.e. a conjunction of a statement and its negation) is true. We could put this symbolically as: p! (q ~q)
15 Intro to Reductio ad Absurdum (RAA) Proofs Now let s say that a statement entails that a contradiction (i.e. a conjunction of a statement and its negation) is true. We could put this symbolically as: p! (q ~q) Well a contradiction by nature is always false (whatever the truth-value of q is, the conjunction will always be false). This is just to say that the negation of a contradiction will always be true. We can capture this fact symbolically as:
16 Intro to Reductio ad Absurdum (RAA) Proofs Now let s say that a statement entails that a contradiction (i.e. a conjunction of a statement and its negation) is true. We could put this symbolically as: p! (q ~q) Well a contradiction by nature is always false (whatever the truth-value of q is, the conjunction will always be false). This is just to say that the negation of a contradiction will always be true. We can capture this fact symbolically as: ~(q ~q)
17 Intro to Reductio ad Absurdum (RAA) Proofs Now let s say that a statement entails that a contradiction (i.e. a conjunction of a statement and its negation) is true. We could put this symbolically as: p! (q ~q) Well a contradiction by nature is always false (whatever the truth-value of q is, the conjunction will always be false). This is just to say that the negation of a contradiction will always be true. We can capture this fact symbolically as: ~(q ~q) Given MT, we can conclude ~p (i.e. p is false). Hence, any statement that entails a contradiction is false.
18 Intro to Reductio ad Absurdum (RAA) Proofs So, say an argument has the following conclusion: p. In order to show p, we can assume ~p (on a line) and seek to derive a contradiction: q! ~q (on a line). The contradiction shows that ~p must be false, which is to say that p must be true! If the conclusion is ~p, we can assume p and seek to derive a contradiction. The contradiction shows that p must be false, which is to say that ~p must be true.
19 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. (F # ~F)! G " G
20 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. (F # ~F)! G " G 2. ~G Assume for RAA
21 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. (F # ~F)! G " G 2. ~G Assume for RAA 3. ~(F # ~F) 1, 2 MT
22 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. (F # ~F)! G " G 2. ~G Assume for RAA 3. ~(F # ~F) 1, 2 MT 4. ~F ~~F 3, DeM
23 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. (F # ~F)! G " G 2. ~G Assume for RAA 3. ~(F # ~F) 1, 2 MT 4. ~F ~~F 3, DeM 5. G 2-4 RAA
24 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. (F # ~F)! G " G 2. ~G Assume for RAA 3. ~(F # ~F) 1, 2 MT 4. ~F ~~F 3, DeM 5. G 2-4 RAA
25 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P
26 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P 3. P Assume for RAA
27 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P 3. P Assume for RAA 4. R Q 1,3 MP
28 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P 3. P Assume for RAA 4. R Q 1,3 MP 5. T ~R 2,3 MP
29 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P 3. P Assume for RAA 4. R Q 1,3 MP 5. T ~R 2,3 MP 6. R 4, Simp
30 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P 3. P Assume for RAA 4. R Q 1,3 MP 5. T ~R 2,3 MP 6. R 4, Simp 7. ~R 5, Simp
31 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P 3. P Assume for RAA 4. R Q 1,3 MP 5. T ~R 2,3 MP 6. R 4, Simp 7. ~R 5, Simp 8. R ~R 6,7 Conj
32 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P 3. P Assume for RAA 4. R Q 1,3 MP 5. T ~R 2,3 MP 6. R 4, Simp 7. ~R 5, Simp 8. R ~R 6,7 Conj 9. ~P 3-8 RAA
33 Intro to Reductio ad Absurdum (RAA) Proofs An Example: 1. P! (R Q) 2. P! (T ~R) " ~P 3. P Assume for RAA 4. R Q 1,3 MP 5. T ~R 2,3 MP 6. R 4, Simp 7. ~R 5, Simp 8. R ~R 6,7 Conj 9. ~P 3-8 RAA
34 Answers to HW We ll come back to RAA on Thursday, for now let s return to our latest predicate logic homework. Ex 9.3 pgs Part B (1-15 All).
35 #s 1-5 # 1. Line 2 is incorrect. (Can t apply UI to a part of a line) #2. Line 5 is incorrect. (Can t univ. generalize a constant that is introduced by EI). Line 8 is incorrect. (Can t existentially instantiate to a variable that occurs earlier in the proof) #3. Line 2 is incorrect. (Can t apply EI to only part of a line) #4. Line 2 is incorrect. (Can t simplify from a existentially quant. Statement) #5. Line 2 is incorrect. (Always instantiate to a constant)
36 #s 6-11 #6. No incorrect moves #7. Line 2 is incorrect. (Need to instantiate to a constant) #8. Line 2 is incorrect. (Can t apply UI to only part of line) #9. Line 3 is incorrect. (Can t apply EG to only part of a line) #10. Line 3 is incorrect. (Line 1 is not a conditional) #11. Line 2 is incorrect. (x in line 1 not uniformly replaced). Line 3 is incorrect. (line 2 not an instance of line 3)
37 #s #12. Line 2 is incorrect. (x in line 1 not uniformly replaced) #13. Line 3 is incorrect. (Line 2 is an instance of line 3. But a cannot appear in line 3) #14. Line 5 is incorrect. (Line 4 is not an instance of line 5) #15. Line 8 is incorrect. (Line 8 is a generalization of a constant that appears in a line derived by EI).
38 More Answers to HW Ex 9.3 pg. 461 Part C (1-8 All):
39 A Few General Tips If the conclusion is an existentially quantified statement, derive an instance of it and apply EG to the instance. If the conclusion is a conditional (or contains a conditional) -- and you re not using CP -- you may need to derive the conditional s corresponding disjunction and then apply MI. The same strategy applies if the conclusion is a disjunction (or contains a disjunction).
40 #1 1. (x)(fx! ~Gx) 2. Fa " ($x)~gx
41 #1 1. (x)(fx! ~Gx) 2. Fa " ($x)~gx 3. Fa! ~Ga 1, UI
42 #1 1. (x)(fx! ~Gx) 2. Fa " ($x)~gx 3. Fa! ~Ga 1, UI 4. ~Ga 2,3 MP
43 #1 1. (x)(fx! ~Gx) 2. Fa " ($x)~gx 3. Fa! ~Ga 1, UI 4. ~Ga 2,3 MP 5. ($x)~gx 4, EG
44 #2 1. Hc # Jd 2. (x)(hx! ($y)ky) 3. (x)(jx! ($y)ly) " ($y)ky # ($y)ly
45 #2 1. Hc # Jd 2. (x)(hx! ($y)ky) 3. (x)(jx! ($y)ly) " ($y)ky # ($y)ly 4. Hc! ($y)ky 2, UI
46 #2 1. Hc # Jd 2. (x)(hx! ($y)ky) 3. (x)(jx! ($y)ly) " ($y)ky # ($y)ly 4. Hc! ($y)ky 2, UI 5. Jd! ($y)ly 3, UI
47 #2 1. Hc # Jd 2. (x)(hx! ($y)ky) 3. (x)(jx! ($y)ly) " ($y)ky # ($y)ly 4. Hc! ($y)ky 2, UI 5. Jd! ($y)ly 3, UI 6. ($y)ky # ($y)ly 1,4,5 CD
48 #3 1. (x)(mx! Ox) 2. ~(Nc # ~Md) " ($x)ox
49 #3 1. (x)(mx! Ox) 2. ~(Nc # ~Md) " ($x)ox 3. ~Nc ~~Md 2, DeM
50 #3 1. (x)(mx! Ox) 2. ~(Nc # ~Md) " ($x)ox 3. ~Nc ~~Md 2, DeM 4. ~~Md 3, Simp
51 #3 1. (x)(mx! Ox) 2. ~(Nc # ~Md) " ($x)ox 3. ~Nc ~~Md 2, DeM 4. ~~Md 3, Simp 5. Md 4, DN
52 #3 1. (x)(mx! Ox) 2. ~(Nc # ~Md) " ($x)ox 3. ~Nc ~~Md 2, DeM 4. ~~Md 3, Simp 5. Md 4, DN 6. Md! Od 1, UI
53 #3 1. (x)(mx! Ox) 2. ~(Nc # ~Md) " ($x)ox 3. ~Nc ~~Md 2, DeM 4. ~~Md 3, Simp 5. Md 4, DN 6. Md! Od 1, UI 7. Od 5,6 MP
54 #3 1. (x)(mx! Ox) 2. ~(Nc # ~Md) " ($x)ox 3. ~Nc ~~Md 2, DeM 4. ~~Md 3, Simp 5. Md 4, DN 6. Md! Od 1, UI 7. Od 5,6 MP 8. ($x)ox 7, EG
55 #4 1. (z)hz 2. (x)(gx! ~Hx) " ($y)(gy! (x)kx)
56 #4 1. (z)hz 2. (x)(gx! ~Hx) " ($y)(gy! (x)kx) 3. Ha 1, UI
57 #4 1. (z)hz 2. (x)(gx! ~Hx) " ($y)(gy! (x)kx) 3. Ha 1, UI 4. Ga! ~Ha 2, UI
58 #4 1. (z)hz 2. (x)(gx! ~Hx) " ($y)(gy! (x)kx) 3. Ha 1, UI 4. Ga! ~Ha 2, UI 5. ~~Ha 3, DN
59 #4 1. (z)hz 2. (x)(gx! ~Hx) " ($y)(gy! (x)kx) 3. Ha 1, UI 4. Ga! ~Ha 2, UI 5. ~~Ha 3, DN 6. ~Ga 4,5 MT
60 #4 1. (z)hz 2. (x)(gx! ~Hx) " ($y)(gy! (x)kx) 3. Ha 1, UI 4. Ga! ~Ha 2, UI 5. ~~Ha 3, DN 6. ~Ga 4,5 MT 7. ~Ga # (x)kx 6, Add
61 #4 1. (z)hz 2. (x)(gx! ~Hx) " ($y)(gy! (x)kx) 3. Ha 1, UI 4. Ga! ~Ha 2, UI 5. ~~Ha 3, DN 6. ~Ga 4,5 MT 7. ~Ga # (x)kx 6, Add 8. Ga! (x)kx 7, MI
62 #4 1. (z)hz 2. (x)(gx! ~Hx) " ($y)(gy! (x)kx) 3. Ha 1, UI 4. Ga! ~Ha 2, UI 5. ~~Ha 3, DN 6. ~Ga 4,5 MT 7. ~Ga # (x)kx 6, Add 8. Ga! (x)kx 7, MI 9. ($y)(gy! (x)kx) 8, EG
63 #5 1. (x)(fx! Gx) 2. (x)(hx! Jx) 3. Fa # Ha " (z)(fz # Hz)! ($x)(gx # Jx)
64 #5 1. (x)(fx! Gx) 2. (x)(hx! Jx) 3. Fa # Ha " (z)(fz # Hz)! ($x)(gx # Jx) 4. Fa! Ga 1, UI
65 #5 1. (x)(fx! Gx) 2. (x)(hx! Jx) 3. Fa # Ha " (z)(fz # Hz)! ($x)(gx # Jx) 4. Fa! Ga 1, UI 5. Ha! Ja 2, UI
66 #5 1. (x)(fx! Gx) 2. (x)(hx! Jx) 3. Fa # Ha " (z)(fz # Hz)! ($x)(gx # Jx) 4. Fa! Ga 1, UI 5. Ha! Ja 2, UI 6. Ga # Ja 3,4,5 CD
67 #5 1. (x)(fx! Gx) 2. (x)(hx! Jx) 3. Fa # Ha " (z)(fz # Hz)! ($x)(gx # Jx) 4. Fa! Ga 1, UI 5. Ha! Ja 2, UI 6. Ga # Ja 3,4,5 CD 7. ($x)(gx # Jx) 6, EG
68 #5 1. (x)(fx! Gx) 2. (x)(hx! Jx) 3. Fa # Ha " (z)(fz # Hz)! ($x)(gx # Jx) 4. Fa! Ga 1, UI 5. Ha! Ja 2, UI 6. Ga # Ja 3,4,5 CD 7. ($x)(gx # Jx) 6, EG 8. ~(z)(fz # Hz) # ($x)(gx # Jx) 7, Add
69 #5 1. (x)(fx! Gx) 2. (x)(hx! Jx) 3. Fa # Ha " (z)(fz # Hz)! ($x)(gx # Jx) 4. Fa! Ga 1, UI 5. Ha! Ja 2, UI 6. Ga # Ja 3,4,5 CD 7. ($x)(gx # Jx) 6, EG 8. ~(z)(fz # Hz) # ($x)(gx # Jx) 7, Add 9. (z)(fz # Hz)! ($x)(gx # Jx) 8, MI
70 #6 1. Ra! Sa 2. (x)(sx! Tx) " ($y)(ry! Ty) 3. Sa! Ta 2, UI 4. Ra! Ta 1, 3 HS 5. ($y)(ry! Ty) 4, EG
71 #7 1. (x)(y)[lx! (Mx Ny)] 2. ~Mb " Lb! Ob 3. (y)[lb! (Mb Ny)] 1, UI 4. Lb! (Mb Nb) 3, UI 5. ~Mb # ~Nb 2, Add 6. ~(Mb Nb) 5, DeM 7. ~Lb 4,6 MT 8. ~Lb # Ob 7, Add 9. Lb! Ob 8, MI
72 #8 1. ~~(v)fv # ~(w)gw 2. ~[($x)hx # ~(w)gw] 3. [(v)fv ~($x)hx]! ~(y)jy 4. Ka! (y)jy " ($z)(kz! Lz) 5. ~($x)hx ~~(w)gw 2, DeM 6. ~~(w)gw 5, Simp 7. ~~(v)fv 1,6 DS 8. (v)fv 7, DN 9. ~($x)hx 5, Simp 10. (v)fv ~($x)hx 8,9 Conj 11. ~(y)jy 3, 10 MP 12. ~Ka 4, 11 MT 13. ~Ka # La 12, Add 14. Ka! La 13, MI 15. ($z)(kz! Lz) 14, EG
73 Ex 9.3 pg. 462 Part D (10-20 Even):
74 Some More Tips -- Always employ EI before UI -- If the conclusion is a universally quantified statement, derive an instance of it and apply UG to the instance.
75 #10 1. (y)(~py! ~Ly) 2. Lc # Ld " Pd # Pc 3. ~Pc! ~Lc 1, UI 4. Lc! Pc 3, Cont 5. ~Pd! ~Ld 1, UI 6. Ld! Pd 5, Cont 7. Pc # Pd 2,4,6 CD 8. Pd # Pc 7, Com
76 #12 1. (z)[uz! (Kz # Sz)] 2. (z)uz 3. ($z)~sz " ($z)kz 4. ~Sa 3, EI 5. Ua! (Ka # Sa) 1, UI 6. Ua 2, UI 7. Ka # Sa 5,6 MP 8. Ka 7,4 DS 9. ($z)kz 8, EG
77 #14 1. (x)[cx! (Dx ($y)ey)] 2. ~Db " ($x)~cx 3. Cb! (Db ($y)ey) 1, UI 4. ~Db # ~($y)ey 2, Add 5. ~(Db ($y)ey) 4, DeM 6. ~Cb 3,5 MT 7. ($x)~cx 6, EG
78 #16 1. (z)~[~(x)jx # ~Kz] " Jc Kc 2. ~[~(x)jx # ~Kc] 1, UI 3. ~~(x)jx ~~Kc 2, DeM 4. (x)jx Kc 3, DN DN 5. (x)jx 4, Simp 6. Jc 5, UI 7. Kc 4, Simp 8. Jc Kc 6,7 Conj
79 #18 1. (x)[(bx! (z)az)! ~P] 2. ($x)~bx " ~P 3. ~Ba 2, EI 4. (Ba! (z)az)! ~P 1, UI 5. (~Ba # (z)az)! ~P 4, MI 6. ~Ba # (z)az 3, Add 7. ~P 5,6 MP
80 #20 1. (x)[(sx ~(z)rz)! Nd] 2. (x)~nx 3. ($x)sx " Rc 4. Sa 3, EI 5. (Sa ~(z)rz)! Nd 1, UI 6. ~Nd 2, UI 7. ~(Sa ~(z)rz) 5,6 MT 8. ~Sa # ~~(z)rz 7, DeM 9. ~~Sa 4, DN 10. ~~(z)rz 8,9 DS 11. (z)rz 10, DN 12. Rc 11, UI
81 Ex 9.3 pgs Part E (5-15 All)
82 #5 1. (x)(jx! ~Ex) 2. ( $x)(jx # Jd) " ($y)(ey! ~Ed) 3. Ja # Jd 2, EI 4. Ja! ~Ea 1, UI 5. Jd! ~Ed 1, UI 6. ~Ea # ~Ed 3,4,5 CD 7. Ea! ~Ed 6, MI 8. ($y)(ey! ~Ed) 7, EG
83 #6 1. (x)(lx! Mx)! (x)(nx! Lx) 2. (x)~lx " (x)~nx 3. ~La 2, UI 4. ~La # Ma 3, Add 5. La! Ma 4, MI 6. (x)(lx! Mx) 5, UG 7. (x)(nx! Lx) 1, 6 MP 8. Na! La 7, UI 9. ~Na 8,3 MT 10. (x)~nx 9, UG
84 #7 1. (x)(sx! Tx) 2. ($y)(ry ~Ty) " ($z)(rz ~Sz) 3. Ra ~Ta 2, EI 4. Sa! Ta 1, UI 5. ~Ta 3, Simp 6. ~Sa 4,5 MT 7. Ra 3, Simp 8. Ra ~Sa 7,6 Conj 9. ($z)(rz ~Sz) 8, EG
85 #8 1. (x)(bx! Cx) 2. (x)(ax! Bx) 3. (x)(cx! Dx) 4. ($x)~dx "($x)~ax 5. ~Da 4, EI 6. Ca! Da 3, UI 7. Aa! Ba 2, UI 8. Ba! Ca 1, UI 9. ~Ca 6,5 MT 10. ~Ba 8,9 MT 11. ~Aa 7, 10 MT 12. ($x)~ax 11, EG
86 #9 1. (x)(rx % Sx) " (x)(rx! Sx) (x)(sx! Rx) 2. Ra % Sa 1, UI 3. (Ra! Sa) (Sa! Ra) 2, ME 4. (Ra! Sa) 3, Simp 5. (x)(rx! Sx) 4, UG 6. (Sa! Ra) 3, Simp 7. (x)(sx! Rx) 6, UG 8. (x)(rx! Sx) (x)(sx! Rx) 5,7 Conj
87 #10 1. (x)[(bx # Ax) % Cx] 2. (x) ~Cx " (x)(ax % Bx) 3. (Ba # Aa) % Ca 1, UI 4. [(Ba # Aa)! Ca] [Ca! (Ba # Aa)] 3, ME 5. ~Ca 2, UI 6. (Ba # Aa)! Ca 4, Simp 7. ~(Ba # Aa) 5,6 MT 8. ~Ba ~Aa 7, DeM 9. ~Aa 8, Simp 10. ~Aa # Ba 9, Add 11. Aa! Ba 10, MI 12. ~Ba 8, Simp 13. ~Ba # Aa 12, Add 14. Ba! Aa 13, MI 15. (Aa! Ba) (Ba! Aa) 11, 14 Conj 16. (Aa % Ba) 15, ME 17. (x)(ax % Bx) 16, UG
88 #11 1. (x)(dx! ~Kx) 2. ($x)(ex Hx) 3. (x)(hx! Dx) 4. (x)(jx! Kx) " ($x)(ex ~Jx) 5. Ea Ha 2, EI 6. Ha! Da 3, UI 7. Ha 5, Simp 8. Da 6,7 MP 9. Da! ~Ka 1, UI 10. ~Ka 8,9 MP 11. Ja!Ka 4, UI 12. ~Ja 10,11 MT 13. Ea 5, Simp 14. Ea ~Ja 13, 12 Conj 15. ($x)(ex ~Jx) 14, EG
89 #12 1. (x)[fx % (Hx ~(y)gy)] 2. ($x)~fx 3. (z)hz " Gc 4. ~Fa 2, EI 5. Fa %(Ha ~(y)gy) 1,UI 6. [Fa!(Ha ~(y)gy)] [(Ha ~(y)gy)! Fa] 5, ME 7. (Ha ~(y)gy)! Fa 6, Simp 8. ~(Ha ~(y)gy) 4,7 MT 9. ~Ha # ~~(y)gy 8 DeM 10. ~~Ha 3 UI, DN 11. (y)gy 9, 10 DS, DN 12. Gc 11, UI
90 #13 1. (x)[bx! (Cx Dx)] 2. ($x)bx " ($x)~(~cx # ~Dx) 3. Ba 2, EI 4. Ba! (Ca Da) 2, UI 5. Ca Da 3,4 MP 6. ~~Ca ~~Da 5, DN DN 7. ~(~Ca # ~Da) 6, DeM 8. ($x)~(~cx # ~Dx) 7, EG
91 #14 1. (x)[mx! ($y)(ny Px)] 2. (x)(nx! ~G) 3. ($x)mx " ~G 4. Ma 3, EI 5. Ma! ($y)(ny Pa) I, UI 6. ($y)(ny Pa) 4,5 MP 7. Nb Pa 6, EI 8. Nb 7, Simp 9. Nb! ~G 2, UI 10. ~G 8,9 MP
92 #15 1. (x)[rx! (Sx # (y)ty)] 2. (x)(rx! Sx)! Pb 3. ~(y)ty " Pb 4. Ra! (Sa # (y)ty) 1, UI 5. ~Ra # (Sa # (y)ty) 4, MI 6. (~Ra # Sa) # (y)ty 5, As 7. ~Ra # Sa 3,6 DS 8. Ra! Sa 7 MI 9. (x)(rx! Sx) 8, UG 10. Pb 2,9 MP
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