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C. Suggested Answers 1. If a student is caught cheating then Alice is angry. 2. Burt and Alice are good writers, and Alice is a good logician but Burt is not. 3. All students who study hard will pass, but some are not happy. 4. No happy students are good writers, but not all students who are good writers are

angry. 5. If all students studied hard, then none will be caught cheating. 6. No students who studied hard are caught cheating. 7. An angry student will be a good writer but will neither study hard nor be good

logician. VIII. CHAPTER EIGHT: PREDICATE LOGIC SEMANTICS A. General Theory 1. If there are no unicorns, what is the truth value of the sentence (x)(Ux ⊃ Mx), where

Ux = x is a unicorn, and Mx = x is mortal? 2. What about the truth value of the sentence (∃x)(Ux ⊃ Mx)? A. Answers 1. The sentence (x)(Ux ⊃ Mx) will be true. To see this, consider the expansion of this sentence for larger and larger universes of discourse. First, in a two−individual universe of discourse, its expansion is (Ua ⊃ Ma) ⋅ (Ub ⊃ Mb), which will be true since each conditional will have a false antecedent. Then for larger universes, each new conjunction added will have a false antecedent and thus be true (on the assumption nothing is a unicorn). 2. The sentence (∃x) (Ux ⊃ Mx) will be true also. Again, consider its expansion for a two−individual universe, namely (Ua ⊃ Ma) ∨ (Ub ⊃ Mb), which will be true because each conditional will have a false antecedent. And obviously, it will be true in larger universes, since adding more disjunctions to a true sentence can't make it false. B. Proving invalidity. Prove that the following arguments are invalid by either the interpretation method or by the expansion method: (1) 1. (∃x)(Ax ⋅ Bx) (3) 1. (x)(~ Ax ⊃ Bx) 2. (∃x) ~ Ax 2. ~ (∃x) ~ Ax /∴(∃x) ~ Bx /∴ ~ (∃x)Bx (2) 1. (x)[(Ax ∨ Bx) ⊃ Cx] (4) 1. ~ (∃x)(~ Ax ⋅ Bx) 2. ~ (Ba ∨ Cb) 2. (∃x)(Ax ⋅ Cx) 3. ~ (Ca ∨ Ab) /∴ ~ (x)(Cx ⊃ ~ Bx)

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/∴~ (~ Aa ∨ ~ Ba) (5) 1. (x)[(Ax ⋅ Bx) ⊃ Cx] /∴ (x)[(Ax ∨ Bx) ⊃ Cx] B. Suggested Answers (1) Let domain be unrestricted, Ax = x is tall, and Bx = x is identical with itself. (2) In a two−individual universe of discourse, the argument amounts to 1. [(Aa ∨ Ba) ⊃ Ca] ⋅ [(Ab ∨ Bb) ⊃ Cb] 2. ~ (Ba ∨ Cb) 3. ~ (Ca ∨ Ab) /∴ ~ (~ Aa ∨ ~ Ba) Let Aa, Ba, Ca, Ab, Bb, and Cb all be false; the premises are true, the conclusion

false. (3) Let the domain of discourse be restricted to the positive integers, and let Ax = x is

an integer, Bx = x is odd. Then we get 1. All positive integers that are not integers are odd. (True, antecedent always F) 2. It is not the case that there is a positive integer that is not an integer. (True) /∴ It is not the case that some positive integer is odd. (False) (4) Restrict the domain of discourse to two individuals, a and b. Then the expansion

for the argument in question will be 1. ~ [(~ Aa ⋅ Ba) ∨ (~ Ab ⋅ Bb)] 2. (Aa ⋅ Ca) ∨ (Ab ⋅ Cb) /∴ ~ [(Ca ⊃ ~ Ba) ⋅ (Cb ⊃ ~ Bb)] This argument has true premises and a false conclusion if Aa and Ca are true and

Ba, Bb, Ab, and Cb are false. (5) Let the domain of discourse be restricted to the positive integers, and let Ax = x >

10; Bx = x > 5; and Cx = x > 7. Then we get 1. (x){[(x > 10) ⋅ (x > 5)] ⊃ (x > 7)} /∴ (x){[(x > 10) ∨ (x > 5)] ⊃ (x > 7)} which has a true premise and a false conclusion. C. Proving consistency. Show that the premises of the arguments below are consistent.

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(1) 1. (x)[(Ax ⋅ Bx) ⊃ Cx] 2. (∃y)(Ay ⋅ ~ Cy) 3. (∃z)(Bz ⋅ ~ Cz) /∴ (∃x)(~ Ax ⋅ ~ Bx) (2) 1. (x)[(Ax ⋅ Bx) ⊃ Cx] 2. Aa ⋅ Ba 3. ~ Cb /∴ ~ Bb C. Suggested Answers (1) Let the domain be the positive integers, Ax = x > 1, Bx = x is odd, Cx = x is > 2: 1. All positive integers that are greater than one and odd are greater than 2. 2. Some positive integer greater than one is not greater than two. 3. Some odd positive integer is not greater than two. (2) Let the domain and predicates be as in the above example, and let a = 3, b = 2: 1. All positive integers that are greater than one and odd are greater than 2. 2. Three is greater than one and odd. 3. Two is not greater than two. IX. CHAPTER NINE: PREDICATE LOGIC PROOFS. Prove valid: (1) 1. ~ (∃x) Ax (5) 1. (∃x)[Ax ⊃ (y)By] /∴ (x)(Ax ⊃ Bx) /∴ (x)Ax ⊃ (y)By (2) 1. ~ Aa (6) 1. (x)Ax ⊃ (y)By /∴ ~ (x) (Ax ⋅ Bx) /∴ (∃x) [Ax ⊃ (y)By] (3) 1. (∃x) (Ax ⋅ Bx) (7) 1. (x)[(Ax ∨ Bx) ⊃ Cx] 2. (x) [Ax ⊃ (Cx ⋅ Dx)] 2. (∃y) ~ (~ Ay ∨ Cy) /∴ (∃x) (Cx ⋅ Dx) /∴ (∃x)(Ax ⋅ Bx) (4) 1. ~ (∃x) ~ (~ Ax ∨ Bx) (8) 1. (x)(Ax ⊃ Bx) ⊃ ~ (∃y) Cy 2. ~ (x)Bx /∴ ~ (x)Ax 2. ~ (x) ~ Cx /∴ (∃x)(Cx ⊃ Ax) IX. SUGGESTED ANSWERS

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(1) 1. ~ (∃x) Ax p 2. (x)~ Ax 1 QN 3. ~ Ax 2 UI 4. ~ Ax ∨ Bx 3 Add 5. Ax ⊃ Bx 4 Impl 6. (x)(Ax ⊃ Bx) 5 UG (2) 1. ~ Aa p 2. ~ Aa ∨ ~ Ba 1 Add 3. ~ (Aa ⋅ Ba) 2 DeM 4. (∃x) ~ (Ax ⋅ Bx) 3 EG 5. ~ (x)(Ax ⋅ Bx) 4 QN (3) 1. (∃x)(Ax ⋅ Bx) p 2. (x)[Ax ⊃ (Cx ⋅ Dx)] p 3. Ax ⋅ Bx 1 EI 4. Ax 3 Simp 5. Ax ⊃ (Cx ⋅ Dx) 2 UI 6. Cx ⋅ Dx 4,5 MP 7. (∃x)(Cx ⋅ Dx) 6 EG (4) 1. ~ (∃x) ~ (~ Ax ∨ Bx) p 2. ~ (x)Bx p 3. ~ ~ (x)Ax AP/∴~ (x)Ax 4. (∃x) ~ Bx 2 QN 5. ~ Bx 4 EI 6. (x)Ax 3 DN 7. Ax 6 UI 8. (x)(~ Ax ∨ Bx) 1 QN 9. ~ Ax ∨ Bx 9 UI 10. ~ ~ Ax 7 DN 11. Bx 9,10 DS 12. Bx ⋅ ~ Bx 5,11Conj 13. ~ (x)Ax 3-12 IP

(5) 1. (∃x)[Ax ⊃ (y)By] p 2. (x)Ax AP/∴(y)By 3. Ax ⊃ (y)By 1 EI 4. Ax 2 UI 5. (y)By 3,4 MP 6. (x)Ax ⊃ (y)By 2-5 CP

(6) 1. (x)Ax ⊃ (y)By p 2. ~ (y)By AP/∴ ~ Ax 3. ~ (x)Ax 1,2 MT 4. (∃x) ~ Ax 3 QN 5. ~ Ax 4 EI 6. ~ (y)By ⊃ ~ Ax 2-5 CP 7. Ax ⊃ (y)By 6 Contra 8. (∃x)[Ax ⊃ (y)By] 7 EG (7) 1. (x)[(Ax ∨ Bx) ⊃ Cx] p 2. (∃y) ~ (~ Ay ∨ Cy) p 3. ~ (∃x)(Ax · Bx) AP… 4. ~ (~ Ay ∨ Cy) 2 EI 5. ~ ~ Ay ⋅ ~ Cy 4 DeM 6. Ay ⋅ ~ Cy 5 DN 7. (Ay ∨ By) ⊃ Cy 1 UI 8. Ay 6 Simp 9. Ay ∨ By 9 Add 10. Cy 7, 9 MP 11. ~ Cy 6 Simp 12. Cy ⋅ ~ Cy 10,11 Conj 13. (∃x)(Ax · Bx) 3-12 IP (8) 1. (x)(Ax ⊃ Bx) ⊃ ~ (∃y)Cy p 2. ~ (x) ~ Cx p 3. (∃x)Cx 2 QN 4. Cx 3 EI 5. (∃y) Cy 4 EG 6. ~ ~ (∃y) Cy 5 DN 7. ~ (x)(Ax ⊃ Bx) 1,6 MT 8. (∃x) ~ (Ax ⊃ Bx) 7 QN 9. ~ (Ay ⊃ By) 8 EI 10. ~ (~ Ay ∨ By) 9 Impl 11. ~ ~Ay ⋅ ~ By) 10 DeM 12. ~ ~Ay 11 Simp 13. Ay 12 DN 14. Ay ∨ ~ Cy 13 Add 15. ~ Cy ∨ Ay 14 Comm 16. Cy ⊃ Ay 15 Impl 17. (∃x)(Cx ⊃ Ax) 16 EG

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X. CHAPTER TEN: RELATIONAL PREDICATE LOGIC

A. Symbolizing 1. If something is heavy, then Art won't lift it. (Hx = “x is heavy”; a = “Art”; Lxy =

“x will lift y”.) 2. If something is heavy, then anyone who is weak will not lift it. (Px = “x is a

person”; Wx = “x is weak”.) 3. Everyone who loves himself hates someone (or other). (Px = “x is a person”; Lxy

= “x loves y”; Hxy = “x hates y”.) 4. All who have sinned have come short of the glory of God. (Px = “x is a person”;

Sx = “x has sinned”; Cxy = “x has come short of y”; g = “the glory of God.”) 5. If anything is heavy, then there is something heavier than this piece of chalk,

which happens not to be heavy. (Hx = “x is heavy”; Hxy = “x is heavier than y”; c = “this piece of chalk”.)

6. All freshmen date only seniors. (Fx = “x is a freshman”; Dxy = “x dates y”; Sx =

“x is a senior”.) 7. Some freshmen date only seniors. 8. Some freshmen date every senior. 9. Some freshmen date no seniors. 10. No freshmen date all seniors. 11. No freshmen date any seniors. 12. Not all freshmen date seniors. 13. Not all freshmen date every senior. 14. Jones will listen to Smith, even though Smith is boring. (j = “Jones”; s = “Smith”;

Lxy = “x will listen to y”; Bx = “x is boring”.) 15. All sinners who sin against someone who is a sinner are sinned against by

someone who is a sinner. (Sx = “x is a sinner”; Sxy = “x sins against y”.) 16. If every sinner is sinned against by everyone who is a sinner, then everyone who

sins against someone (or other) sins against himself.

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17. Someone who is a sinner and sins against someone (or other) who is a sinner sins against himself.

18. All heads of horses are heads of animals, provided that all horses are animals. (Hx

= “x is a horse”; Ax = “x is an animal”; Hxy = “x is the head of y”.) A. Answers 1. (x) (Hx ⊃ ~ Lax) 2. (x) {Hx ⊃ (y)[(Py ⋅ Wy) ⊃ ~ Lyx]} 3. (x) [(Px ⋅ Lxx) ⊃ (∃y)(Py ⋅ Hxy)] 4. (x) [(Px ⋅ Sx) ⊃ Cxg] 5. (∃x) Hx ⊃ (∃x) (Hxc ⋅ ~ Hc) 6. (x) [Fx ⊃ (y) (Dxy ⊃ Sy)] 7. (∃x) [Fx ⋅ (y) (Dxy ⊃ Sy)] 8. (∃x) [Fx ⋅ (y) (Sy ⊃ Dxy)] 9. (∃x) [Fx ⋅ (y) (Sy ⊃ ~ Dxy)] or (∃x) [Fx ⋅ ~ (∃y) (Sy ⋅Dxy)] 10. (x) [Fx ⊃ ~ (y) (Sy ⊃ Dxy)] or ~ (∃x) [Fx ⋅ (y) (Sy ⊃ Dxy)] 11. (x) [Fx ⊃ (y) (Sy ⊃ ~ Dxy )] or ~ (∃x) [Fx ⋅ (∃y) (Sy ⋅ Dxy)] 12. ~ (x) [Fx ⊃ (∃y) (Sy ⋅ Dxy)] 13. ~ (x) [Fx ⊃ (y) (Sy ⊃ Dxy )] 14. Ljs ⋅ Bs 15. (x) {[Sx ⋅ (∃y) (Sy ⋅ Sxy)] ⊃ (∃z) (Sz ⋅ Szx)} 16. (x) [Sx ⊃ (y) (Sy ⊃ Syx) ] ⊃ (x) [(∃y) Sxy ⊃ Sxx] 17. (x) {[Sx ⋅ (∃y) (Sy ⋅ Sxy)] ⊃ Sxx} 18. (x)(Hx ⊃ Ax) ⊃ (x)[(∃y)(Hy ⋅ Hxy) ⊃ (∃z)(Az ⋅ Hxz)] B. Translations. Letting Hxy = “x is heavier than y”, translate the following into English (coming as close as possible to colloquial usage): 1. (x) (y) Hxy 2. (∃x) (∃y) Hxy 3. (x) (∃y) Hxy 4. (∃x) (y) Hyx 5. (∃x) (y) Hxy

6. ~ (∃x)(y)Hxy 7. (x) [(y) Hxy ⊃ Hxx] 8. (∃x) ~ Hxx ⊃ ~ (∃x) (y) Hxy 9. (x) [(∃y) Hxy ⊃ (∃z) Hzx] 10. (x) [(y) Hyx ⊃ ~ (∃z) Hxz]

B. Answers 1. Everything is heavier than everything. 2. Something is heavier than something. 3. Everything is heavier than something.

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4. There is something such that everything is heavier than it. 5. Something is heavier than everything. 6. Nothing is heavier than everything. Or It's not the case that something is heavier

than everything. 7. Anything heavier than everything is heavier than itself. 8. If there is something that is not heavier than itself, then nothing is heavier than

everything. 9. Everything heavier than something (or other) is such that something is heavier

than it. 10. Anything that everything is heavier than is not heavier than anything. C. Proving invalidity. Show that the following arguments are invalid: (1) 1. (∃x) (∃y) Fxy/∴ Faa (2) 1. (∃x) (∃y) (Fxy ⊃ Gx) /∴ (∃x) ((∃y) Fxy ⊃ Gx) C. Answers (1) 1.(Faa ∨ Fab) ∨ (Fba ∨ Fbb) F T T T F F F /∴ Faa F (2) 1. [(Faa ⊃ Ga) ∨ (Fab ⊃ Ga)] ∨ [(Fba ⊃ Gb) ∨ (Fbb ⊃ Gb)] T F F T F T F T T F F T F T F /∴ [(Faa ∨ Fab) ⊃ Ga] ∨ [(Fba ∨ Fbb) ⊃ Gb] T T F F F F T T F F F D. Prove valid (1) 1. (x) [Ax ⊃ (y) (By ⊃ ~ Cxy)] 2. (x)[Ax ⊃ (y)(Dy ⊃ Cxy )] 3. (∃x) Ax /∴ (∃x) [(y) (Dy ⊃ Cxy) ⋅ (z) (Bz ⊃ ~ Cxz)] (2) 1. (x) (Ax ⊃ Bxa) 2. (x) (y) [(Ax ⋅ Bxy) ⊃ ~ Byx] /∴ (x) (Ax ⊃ ~ Bax) (3) 1. (x)(y) [(Axy ⋅ Bxy) ⊃ (∃z) (Cz ⋅ Dxzy)] 2. Aab ⊃ Bab /∴ ~ Aab ∨ ~ (x)(Cx ⊃ ~ Daxb) (4) 1. (x) (y) [Ay ⋅ (Bx ⊃ Cx)] 2. (x) [(∃y) ~ Dxy ⊃ Bx] /∴ (x) [~ (∃y) (Ay ⋅ Dxy) ⊃ Cx]

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(3) 1. (x)(y)[(Axy ⋅ Bxy) ⊃ (∃z) (Cz ⋅ Dxzy)] p 2. Aab ⊃ Bab p 3. Aab AP/∴ ~ (x)(Cx ⊃ ~ Daxb) 4. Bab 2,3 MP 5. (y)[(Aay ⋅ Bay) ⊃ (∃z) (Cz ⋅ Dazy) 1 UI 6. (Aab ⋅ Bab) ⊃ (∃z) (Cz ⋅ Dazb) 5 UI 7. Aab ⋅ Bab 3,4 Conj 8. (∃z)(Cz ⋅ Dazb) 6,7 MP 9. (Cz ⋅ Dazb) 8 EI 10. ~ ~ (Cz ⋅ Dazb) 9 DN 11. ~ (~ Cz ∨ ~ Dazb) 10 DeM 12. ~ (Cz ⊃ ~ Dazb) 11 Impl 13. (∃x) ~ (Cx ⊃ ~ Daxb) 12 EG 14. ~ (x)(Cx ⊃ ~ Daxb) 13 QN 15. Aab ⊃ ~ (x)(Cx ⊃ ~ Daxb) 3-14 CP 16. ~ Aab ∨ ~ (x)(Cx ⊃ ~ Daxb) 15 Impl (4) 1. (x)(y)[Ay ⋅ (Bx ⊃ Cx)] p 2. (x) [(∃y) ~ Dxy ⊃ Bx] p 3. ~ (∃y) (Ay ⋅ Dxy) AP/∴ Cx 4. (y) ~ (Ay ⋅ Dxy) 3 QN 5. ~ (Ay ⋅ Dxy) 4 UI 6. ~ Ay ∨ ~ Dxy 5 DeM 7. (y) [Ay ⋅ (Bx ⊃ Cx)] 1 UI 8. Ay ⋅ (Bx ⊃ Cx) 7 UI 9. Ay 8 Simp 10. ~ ~ Ay 9 DN 11. ~ Dxy 6,10 DS 12. (∃y) ~ Dxy 11 EG 13. (∃y) ~ Dxy ⊃ Bx 2 UI 14. Bx 12,13 MP 15. Bx ⊃ Cx 8 Simp 16. Cx 14,15 MP 17. ~ (∃y) (Ay ⋅ Dxy) ⊃ Cx 3-16 CP 18. (x) [~ (∃y) (Ay ⋅ Dxy) ⊃ Cx] 17 UG

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E. Theorems of logic. Prove that the following are theorems of logic: 1. (x)Ax ⊃ (∃x)Ax 2. [(x)Ax ⊃ ~ (∃y) By ] ⊃ (∃x)[~ Ax ∨ (y) ~ By] 3. (x)(Axa ⊃ Bax) ⊃ [(∃x)Axa ⊃ (∃x)Bax] 4. (∃x)[(∃y) Aya ⊃ Bax] ≡ [(x)~ Bax ⊃ (y)~ Aya] E. Suggested Answers (1) 1. (x)Ax AP/∴(∃x)Ax 2. Ax 1 UI 3. (∃x)Ax 2 EG 4. (x)Ax ⊃ (∃x)Ax 1-3 CP (2) 1. (x)Ax ⊃ ~ (∃y) By AP/∴ (∃x)[~ Ax ∨ (y) ~ By] 2. ~ (y) ~ By AP/∴ ~ Ax 3. (x)Ax ⊃ (y) ~ By 1 QN 4. ~ (x)Ax 2,3 MT 5. (∃x) ~ Ax 4 QN 6. ~ Ax 5 EI 7. ~ (y) ~ By ⊃ ~ Ax 2-6 CP 8. Ax ⊃ (y) ~ By 7 Contra 9. ~ Ax ∨ (y) ~ By 8 Impl 10. (∃x)[~ Ax ∨ (y) ~ By] 9 EG 11. [(x)Ax ⊃ ~ (∃y ) By] ⊃ (∃x)[~ Ax ∨ (y) ~ By] 1-10 CP (3) 1. (x)(Axa ⊃ Bax) AP/∴ (∃x)Axa ⊃ (∃x)Bax 2. (∃x)Axa AP/∴ (∃x) Bax 3. Axa 2 EI 4. Axa ⊃ Bax 1 UI 5. Bax 3,4 MP 6. (∃x) Bax 5 EG 7. (∃x)Axa ⊃ (∃x)Bax 2-6 CP 8. (x)(Axa ⊃ Bax ) ⊃ [(∃x)Axa ⊃ (∃x)Bax] 1-7 CP

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(4) 1. (∃x)[(∃y) Aya ⊃ Bax] AP/∴ (x) ~ Bax ⊃ (y) ~ Aya 2. (x) ~ Bax AP/∴ (y) ~ Aya 3. (∃y) Aya ⊃ Bax 1 EI 4. ~ Bax 2 UI 5. ~ (∃y) Aya 3,4 MT 6. (y) ~ Aya 5 QN 7. (x) ~ Bax ⊃ (y) ~ Aya 2-6 CP 8. (∃x)[(∃y) Aya ⊃ Bax ] ⊃ [(x) ~ Bax ⊃ (y) ~ Aya] 1-7 CP 9. (x) ~ Bax ⊃ (y) ~ Aya AP/∴ (∃x)[(∃y) Aya ⊃ Bax] 10. ~ (∃x)[(∃y) Aya ⊃ Bax] AP/∴ (∃x)[(∃y) Aya ⊃ Bax] 11. (x) ~ [(∃y ) Aya ⊃ Bax] 10 QN 12. ~ [(∃y ) Aya ⊃ Baz] 11 UI 13. ~ [~ (∃y ) Aya ∨ Baz] 12 Impl 14. ~ ~ (∃y) Aya ⋅ ~ Baz 13 DeM 15. ~ ~ (∃y) Aya 14 Simp 16. ~ (y) ~ Aya 15 QN 17. ~ Baz 14 Simp 18. (x) ~ Bax 17 UG 19. (y) ~ Aya 9,18 MP 20. (y) ~ Aya ⋅ ~ (y) ~ Aya 16,19 Conj 21. (∃x)[(∃y) Aya ⊃ Bax] 10-20 IP 22. [(x) ~ Bax ⊃ (y) ~ Aya] ⊃ (∃x)[(∃y)Aya ⊃ Bax] 9-21 CP 23. 8 ⋅ 22 8,22 Conj 24. (∃x)[(∃y)Aya ⊃ Bax] ≡ [(x) ~ Bax ⊃ (y) ~ Aya] 23 Equiv

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XI. CHAPTER ELEVEN: QUANTIFIER RULES THEORY 1. Explain, using a concrete example, why we should not

1. (∃y) Lxy

allow the following inference when using EI. What restriction then is required on the use of EI?

2. Lxx 1 EI 2. Explain, using a concrete example, why we should not

1. (∃y) Lxy

allow the following inference when using UG. What restriction then is required on the use of UG?

2. Lxy 1 EI 3. (y) Lxy 2 UG 3. Explain, using a concrete example, why we should not

1. Lxy

allow the following inference when using EG. What restriction then is required on the use of EG?

2. (∃x)Lxx 1 EG 4. Carefully explain why we forbid inferences like this one from line 2 to line 4: 1. (∃x)(Rx ⋅ Sx) p 2. (∃y)(By ⋅ Ty) p 3. Rx ⋅ Sx 1 EI 4. Bx ⋅ Tx 2 EI 5. Carefully explain why we forbid inferences like this one from line 1 to line 2: 1. (x) (∃y) (Hx ⊃ Lyx) p

2. (∃y) (Hy ⊃ Lyy) 1 UI 6. Do the same for the inference from line 3 to line 4: 1. (x) (∃y) (Rx ⊃ Byx) p 2. (∃y) (Rx ⊃ Byx) 1 UI 3. Rx ⊃ Byx 2 EI 4. (z) (Rz ⊃ Byz) 3 UG 5. (∃y) (z) (Rz ⊃ Byz) 4 EG 7. Suppose we require that in using UG and EG there should be a one-to-one correspondence between free u’s in (. . .u. . .) and bound w’s in (w) (. . .w. . .) or (∃w) (. . .w. . .). What errors or omissions (incompleteness), if any, would result? 8. Suppose we similarly require that in using UI and EI there should be a one-to-one correspondence between bound w’s in (w) (. . .w. . .) or (∃w) (. . .w. . .) and free u’s in (. . .u. . .). What errors or omissions, if any, would result?

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8.We get into similar trouble with UI. That is, we would block valid inferences using UI, such as: 1. (x) Oxy 2. Oyy 1 UI (valid) For example, if 1 says that everything is as old as itself, then it follows for y that it is as old as itself. For another example, consider the inference below: 1. (∃x) (y) Bxy 2. (y) Bxy 1 EI 3. Bxx 2 UI (valid) 4. (∃x) Bxx 3 EG Suppose the domain limited to people and Bxy = x benefits y. Then 1 says that someone benefits everyone, and if that is true then it follows that there is someone who benefits him or herself, that is, 4.

XII. CHAPTER TWELVE: PREDICATE LOGIC TRUTH TREES

Use the truth-tree method to determine the validity/invalidity of the following arguments. If invalid, use the constants in an open path to expand the premises and conclusion and assign truth-values to show the invalidity. (1) 1. (x) ~ (Ax ⋅ ~ Bx) 2. (∃x) (~ Bx ⋅ Cx) /∴ (∃x) ~ Ax (2) 1. (x) ~ Lx 2. (∃x) Lx ∨ (∃x) Mx /∴ (∃x) Mx (3) 1. ~ (∃x) (Ax ∨ Bx) 2. (x)[(Cx ∨ Dx) ⊃ Ax] /∴ ~ (∃x) Cx (4) 1. (∃y) (x)Lyx /∴ (x) (∃y) Lyx (5) 1. (x) [Fx ⊃ (∃y) Gxy] /∴ (x) (∃y) [Fx ⊃ Gxy]

(6) 1. (x)[Lx ⊃ (Ax ∨ Bx)] 2. (x)[(Mx ⋅ Lx) ⊃ ~ Ax] /∴ (x)(Mx ⊃ Lx) ⊃ (x)(Mx ⊃ Bx) (7) 1. (∃x) (Px ≡ Qx) 2. (∃x) (Px · Rx) /∴ (∃x) (Qx ≡ Rx) (8) 1. (∃x) (∃y) Fxy ⊃ (∃x) (∃y) Gxy 2. (∃x) Fxx /∴ (∃x) Gxx

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XII. ANSWERS (1) 1. (x) ~ (Ax ⋅ ~ Bx) p 2. (∃x) (~ Bx ⋅ Cx) p/∴ (∃x) ~ Ax 3. ~ (∃x) ~ Ax Denial of conclusion 4. (x) ~ ~ Ax 3 FC2 5. ~ Ba ⋅ Ca 2 FC4 6. ~ Ba 5 FC3 Ca 7. ~ (Aa ⋅ ~ Ba) 1 FC5 8. ~ ~ Aa 4 FC5 9. Aa 8 FC2 10. ~ Aa ~ ~ Ba 7 FC3 11. 12. Ba 10 FC2 13. Valid. All paths close. (2) 1. (x) ~ Lx p 2. (∃x) Lx ∨ (∃x) Mx p/∴ (∃x) Mx 3. ~ (∃x) Mx Denial of conclusion 4. (x) ~ Mx 3 FC2 5. (∃x) Lx (∃x) Mx 2 FC3 6. La Ma 5 FC4 7. ~ La 1 FC5 ~ Ma 4 FC4 8. Valid. All paths close. (3) 1. ~ (∃x) (Ax ∨ Bx) p 2. (x)[(Cx ∨ Dx) ⊃ Ax] p/∴ ~ (∃x) Cx 3. ~ ~ (∃x) Cx Denial of conclusion 4. (∃x) Cx 3 FC2 5. (x) ~ (Ax ∨ Bx) 1 FC2 6. Ca 4 FC4 7. (Ca ∨ Da) ⊃ Aa 2 FC5 8. ~ (Aa ∨ Ba) 5 FC5 9. ~ Aa 8 FC3 ~ Ba 10. ~ (Ca ∨ Da) Aa 7 FC3 11. 12. ~ Ca 10 FC3 ~ Da 13. Valid. All paths close.

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(4) 1. (∃y) (x)Lyx p/∴ (x) (∃y) Lyx 2. ~ (x) (∃y) Lyx Denial of conclusion 3. (∃x) ~ (y) Lyx 2 FC2 4. (y) ~ Lya 3 FC4 5. (x) Lbx 1 FC4 6. ~ Laa 4 FC5 ~ Lba 7. Lba 5 FC5 Lbb 8. Valid. All paths close. (5) 1. (x) [Fx ⊃ (∃y) Gxy] p/∴ (x) (∃y) (Fx ⊃ Gxy) 2. ~ (x) (∃y) (Fx ⊃ Gxy) Denial of conclusion 3. (∃x) ~ (∃y) (Fx ⊃ Gxy) 2 FC2 4. ~ (∃y) (Fa ⊃ Gay) 3 FC4 5. (y) ~ (Fa ⊃ Gay) 4 FC2 6. Fa ⊃ (∃y) Gay 1 FC5 7. ~ (Fa ⊃ Gaa) 5 FC5 8. Fa 7 FC3 ~ Gaa 9. ~ Fa (∃y) Gay 6 FC3 10. 11. Gab 9 FC4 12. Fb ⊃ (∃y) Gby 1 FC5 13. ~ (Fa ⊃ Gab) 5 FC5 14. Fa 13 FC3 ~ Gab 15. Valid. All paths close.

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(8) 1. (~ p ∨ q) ⋅ (~ r ⊃ ~ q) p 2. s ⋅ (r ∨ p) p 3. ~ (s ⋅ r) Negation of conclusion 4. s From line 2 r ∨ p 5. ~ p ∨ q From line 1 ~ r ⊃ ~ q 6. ~ s ~ r From line 3 7. r p From line 4 8. ~ p q From line 5 9. ≁≁r ~ q From line 5 All paths closed. Valid.

VII. CHAPTER SEVEN: PREDICATE LOGIC SYMBOLIZATION

A. General theory: 1. Which of the following are sentences? a. (x)Fx ⊃ (∃y)Gxy d. ~ (∃y) Fy ⊃ (x)Gx b. (x)Fx ⊃ Ga e. None of these c. (x)(Fx ⊃ Ga) 2. Which of the following are sentences? a. (x)(Fx ⊃ Gx) d. (x)Fy ⊃ (∃y)Gx b. (x)Fx ⊃ (∃x)Gx e. None of these c. (x)Fx ⊃ Gx 3. Symbolize “Some mammals are not four-legged” a) when the domain is mammals

and b) when the domain is unrestricted (using obvious abbreviations). 4. Symbolize “No whales are fish” a) when the domain is whales, and b) when the

domain is unrestricted (using obvious abbreviations).

5. What sentence contradicts “No just acts are acts that cause pain”? a. All just acts are acts that cause pain. b. Some just acts are acts that cause pain. c. Some just acts are acts that do not cause pain. d. No acts that cause pain are just acts. e. None of these.

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6. Which sentence is equivalent to (∃x)(Ax · ~ Bx)? a. (x)(Ax ⊃ ~ Bx) c. ~ (x)(Ax ⊃ Bx) b. (x)(Ax ⊃ Bx) d. None of these 7. Which sentence is equivalent to (∃x)(Ax · Bx)? a. ~ (x)(Ax ⊃ ~ Bx) c. ~ (x)(Ax ⊃ Bx) b. ~ (∃x)(Ax · ~ Bx) d. None of these A. Answers 1. b, c, d 2. a, b 3. a) (∃x) ~ Fx, b) (∃x)(Mx · ~ Fx) 4. a) ~ (∃x)Fx or (x) ~ Fx, b) ~ (∃x)(Wx · Fx) or (x)(Wx ⊃ ~ Fx) 5. b 6. c 7. a B. Symbolize, using the indicated letters: 1. No mathematician philosophers are scientists. (Mx = “x is a mathematician”; Px =

“x is a philosopher”; Sx = “x is a scientist”.) 2. All mathematicians and philosophers are either non−philosophers or

non−scientists. 3. Some mathematicians and (some) philosophers are scientists. 4. No one is a scientist unless he also is a mathematician and philosopher. 5. Only mathematicians are scientists, and none but philosophers are

mathematicians. 6. No arguments that are either invalid or unsound are convincing. (Ax = “x is an

argument”, Vx = “x is valid”; Sx = “x is sound”; Cx = “x is convincing”.) 7. A student caught cheating will be expelled. (Sx = “x is a student”; Cx = “x is

caught cheating”; Ex = “x will be expelled”.) 8. A student who was caught cheating was not expelled. 9. If all students are good logicians, then they all will pass. (Sx = “x is a student”; Gx

= “x is a good logician”; Px = “x will pass”.) 10. If a student is a good logician, then he or she will pass.

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D. Suggested Answers (1) 1. (x)[Ax ⊃ (y)(By ⊃ ~ Cxy)] p 2. (x)[Ax ⊃ (y)(Dy ⊃ Cxy)] p 3. (∃x)Ax p 4. Ax 3 EI 5. Ax ⊃ (y)(By ⊃ ~ Cxy) 1 UI 6. (y)(By ⊃ ~ Cxy) 4,5 MP 7. (Bz ⊃ ~ Cxz) 6 UI 8. (z)(Bz ⊃ ~ Cxz) 7 UG 9. Ax ⊃ (y)(Dy ⊃ Cxy) 2 UI 10. (y)(Dy ⊃ Cxy) 4,9 MP 11. (y)(Dy) ⊃ Cxy) ⋅ (z)(Bz ⊃ ~ Cxz) 8,10Conj 12. (∃x)[(y)(Dy ⊃ Cxy) ⋅ (z)(Bz ⊃ ~ Cxz)] 11 EG (2) 1. (x)(Ax ⊃ Bxa) p 2. (x)(y)[(Ax ⋅ Bxy) ⊃ ~ Byx] p 3. Ax AP/∴ ~ Bax 4. Ax ⊃ Bxa 1 UI 5. Bxa 3,4 MP 6. (y) [(Ax ⋅ Bxy) ⊃ ~ Byx] 2 UI 7. (Ax ⋅ Bxa) ⊃ ~ Bax 6 UI 8. Ax ⋅ Bxa 3,5 Conj 9. ~ Bax 7,8 MP 10. Ax ⊃ ~ Bax 3-9 CP 11. (x)(Ax ⊃ ~ Bax) 10 UG

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XI. ANSWERS 1. It could allow us to move from a true statement to a false one. For example, if Lxy = “x loves y”, then from the fact that x loves someone it does not follow that x loves him or herself. EI requires that the variable replacing the variable made free with the removal of the existential quantifier be one that is not free earlier in the proof (restriction 2). 2. It could allow us to move from a true statement to a false one. For example, if Lxy = “x loves y”, then from the fact that x loves someone it does not follow that x loves everyone. UG requires that the variable that is universally quantified be one that does not occur previously in a line obtained by EI (restriction 2) 3. It could allow us to move from a true statement to a false one. For example, if Lxy = “x loves y”, from the fact that someone loves someone it does not follow that anyone loves him or herself. EG requires that the variable that comes to be existentially quantified be one that does not already occur free in the formula we are quantifying. 4. It could allow us to move from true statements to a false one. In particular, if we allow such inferences, we could prove that one particular item has some properties A and B given the premises that some x has A and some x (not necessarily the same one) has B. Thus, in the example, if one asserts there are short redheads and two that there are tall blondes, we wouldn’t want to conclude to 5. (Rx ⋅ Sx) ⋅ (Bx ⋅ Tx) 3, 4 Conj and 6. (∃x) [(Rx ⋅ Sx) ⋅ (Bx ⋅ Tx)] 5 EG For 6 says that there are some short redheads who are tall blondes—rather unlikely. 5. The point of dropping a quantifier is to free the variables it bound. In this case, the variables “freed” by dropping the (x) quantifier are “snatched up” by the (∃y) quantifier. Suppose l states that every handsome person is loved by someone. Then 2 states that there is someone who, if handsome, loves himself. Clearly 1 could be true when 2 is false. 6. The x's free in line 3 were free in line 2 to which EI was applied; thus free x’s acquire the “taint” of EI even though, as in this case, they were freed by UI. Let line 1 say that every rich person has someone who bows to him. And then line 5 will say that there is someone who bows to all rich people, an entirely different matter. 7. No errors, one omission. We would block valid inferences using EG, such as 1. Oxx 2. (∃y) Oyx 1 EG (valid) For example, if 1 says that x is as old as itself, then it is true that there is something that is as old as x.

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(6) 1. (x)[Lx ⊃ (Ax ∨ Bx)] p 2. (x)[(Mx ⋅ Lx) ⊃ ~ Ax] p/∴(x)(Mx ⊃ Lx) ⊃ (x)(Mx ⊃ Bx) 3. ~ [(x)(Mx ⊃ Lx) ⊃ (x)(Mx ⊃ Bx)] Denial of conclusion 4. (x)(Mx ⊃ Lx) 3 FC3 ~ (x)(Mx ⊃ Bx) 5. (∃x) ~ (Mx ⊃ Bx) 4 FC2 6. ~ (Ma ⊃ Ba) 5 FC4 7. Ma 6 FC3 ~ Ba 8. La ⊃ (Aa ∨ Ba) 1 FC5 9. (Ma ⋅ La) ⊃ ~ Aa 2 FC5 10. Ma ⊃ La 4 FC5 11. ~ Ma La 10 FC3 12. 13. ~ La (Aa ∨ Ba) 8 FC3 14. 15. ~ (Ma ⋅ La) ~ Aa 9 FC3 16. ~ Ma ~ La 15 FC3 Aa Ba 13 FC3 17.

Valid. All paths close. (7) 1. (∃x) (Px ≡ Qx) p 2. (∃x) (Px · Rx) p/∴ (∃x) (Qx ≡ Rx) 3. ~ (∃x) (Qx ≡ Rx) Denial of conclusion 4. (x) ~ (Qx ≡ Rx) 3 FC2 5. Pa · Ra 2 FC4 6. Pa 5 FC3 Ra 7. Pb ≡ Qb 1 FC4 8. Pb ~ Pb 7 FC3 Qb ~ Qb 9. ~ (Qa ≡ Ra) ~ (Qa ≡ Ra) 4 FC5 10. ~ (Qb ≡ Rb) ~ (Qb ≡ Rb) 4 FC5 11. Qa ~ Qa 9 FC3 ~ Ra Ra 12. 13. Qb ~ Qb 10 FC3 ~ Rb Rb

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14. *

Invalid. Open path. In the branch marked with *, let V(Pa) = T, V(Pb) = T, V(Qa) = F, V(Qb) = T, V(Ra) = T, V(Rb) = F. The expansions give us 1. (Pa ≡ Qa) ∨ (Pb ≡ Qb) 2. (Pa ⋅ Ra) ∨ (Pb ⋅ Rb) /∴ (Qa ≡ Ra) ∨ (Qb ≡ Rb), thus true premises, false conclusion. (8) 1. (∃x) (∃y) Fxy ⊃ (∃x) (∃y) Gxy p 2. (∃x) Fxx p/∴ (∃x) Gxx 3. ~ (∃x) Gxx Denial of concl. 4. (x) ~ Gxx 3 FC2 5. ~ (∃x) (∃y) Fxy (∃x) (∃y) Gxy 1 FC3 6. (x) ~ (∃y) Fxy 5 FC2 Faa 2 FC4 7. Faa 2 FC4 (∃y) Gby 5 FC4 8. ~ Gaa 4 FC5 Gbc 7 FC4 9. ~ (∃y) Fay 6 FC5 ~ Gaa 4 FC5 10. (y) ~ Fay 9 FC2 ~ Gbb 11. ~Faa 10 FC5 ~ Gcc 12. * Invalid. Open path. In the branch marked with *, let V(Faa) = T, V(Gbc) = T V(Gaa) = F, V(Gbb) = F V(Gcc) = F (let the values for the other relevant atomic sentences be T or F). The expansions give us 1. ([Faa ∨ (Fab ∨ Fac)] ∨ {[Fba ∨ (Fbb ∨ Fbc)] ∨ [Fca ∨ (Fcb ∨ Fcc)]}) ⊃ ([Gaa ∨ (Gab ∨ Gac)] ∨ {[Gba ∨ (Gbb ∨ Gbc)] ∨ [Gca ∨ (Gcb ∨ Gcc)]}) 2. Faa ∨ (Fbb ∨ Fcc) /∴ Gaa ∨ (Gbb ∨ Gcc), thus true premises, false conclusion.

XIII. CHAPTER THIRTEEN: IDENTITY

A. General Theory 1. Determine the status of the following relations with respect to symmetry,

transitivity, and reflexivity: a) ______________ respects _ _ _ _ _ _ _ _ _ b) ______________ resembles _ _ _ _ _ _ _ _ _ c) ______________ is a child of _ _ _ _ _ _ _ _ _

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2. Explain why translating “s is visible to Jane” as a material conditional, such as “if s is near Jane (and the line of sight is unobstructed, etc.) then it will be seen by her,” seems to be inadequate.

3. Use an example to explain the distinction that is made between sentences and

propositions. 4. Give two reasons why some think that universal statements cannot be adequately

expressed as conjunctions. A. Answers 1. a) nonsymmetrical, nontransitive, nonreflexive b) symmetrical, nontransitive, totally reflexive c) asymmetrical, intransitive, irreflexive 2. The main problem stems from the fact that a false antecedent is sufficient for the

material conditional being true. In this case then something would be visible to Jane just by the fact that it was never near her. But this could be true of something that is clearly not visible to her.

3. Sentences are series of ink marks on paper or sounds in that air, and thus can be

seen or heard. Examples of sentences are “Mary sees John” and “John is seen by Mary.” These are each tokens of different sentence types. These sentences however express the same proposition. So it seems that propositions cannot be indentified with particular sentences, and thus seem to be a different kind of thing.

4. First, because in some cases the domain of discourse is infinite, and thus to give

the equivalent conjunction would require writing down an infinite number of conjuncts. Second, the universal claim says that everything in the domain of discourse has some property, but the expansion does not since it does not involve the claim that the individuals in the expansion are the only individuals there are in the domain of discourse.

B. Proofs with identity. Prove the validity of the following arguments: (1) 1. (x) (Ax ⊃ Bx) 2. ~ Ba 3. a = b /∴ ~ Ab (2) 1. Aab 2. Bc 3. a = c / ∴ (∃x) (Bx ⋅ Axb) (3) 1. (x) ~ (Ax ⋅ Bx)

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2. (x) (Ax ⊃ Bx) 3. Aa ⋅ ~ Bb / ∴ ~ (a = b)

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B. Answers (1) 1. (x) (Ax ⊃ Bx) p 2. ~ Ba p 3. a = b p 4. Aa ⊃ Ba 1 UI 5. ~ Aa 2, 4 MT 6. ~ Ab 3, 5 ID (2) 1. Aab p 2. Bc p 3. a = c p 4. Ba 2, 3 ID 5. Ba ⋅ Aab 1, 4 Conj 6. (∃x) (Bx ⋅ Axb) 5 EG (3) 1. (x) ~ (Ax ⋅ Bx) p 2. (x) (Ax ⊃ Bx) p 3. Aa ⋅ ~ Bb p 4. ~ ~ (a = b) AP /∴~ (a = b) 5. a = b 4 DN 6. Aa 3 Simp 7. Aa ⊃ Ba 2 UI 8. Ba 6, 7 MP 9. ~ Bb 3 Simp 10. Bb 5, 8 ID 11. Bb ⋅ ~ Bb 9, 10 Conj 12. ~ (a = b) 4-11 IP C. Symbolize, using the indicated letters: 1. Something that is heavier than everything else is still not heavier than itself. (Hxy

= “x is heavier than y “.) 2. Anything that is heavier than itself is heavier than something else. 3. Any sinner who sins against someone else sins against everyone else. (Sx = “x is a

sinner”; Sxy = “x sins against y”.) 4. The sinner who sins against every sinner sins against himself. 5. All sinners who sin against everyone else, sin against themselves. 6. One and only one sinner sins against all sinners.

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7. Not all sinners are sinned against by other sinners. 8. No sinner is sinned against by all other sinners. 9. At most one sinner sins against everyone else who is a sinner. (Quite difficult) 10. If some sinners are not sinned against by others, then for any nonsinners, they

aren't sinned against by any sinners. 11. The largest Great Lake is Lake Superior (Gx = “x is a Great Lake”; Lxy = “x is

larger than y”; s = “Lake Superior”) 12. The capitol of Ohio is a river town. (Cxy = “x is a capitol of y”; Rx = “x is a river

town”; o = “Ohio”) 13. The capitol of Ohio is Columbus. (c = “Columbus”) 14. The capitol of Ohio is not the largest city in Ohio. (Ixy = “x is a city in y”) 15. Every state has exactly two senators. (Sx = “x is a state”; Sxy = “x is a senator

from y”) C. Answers 1. (x){(y)[(x ≠ y) ⊃ Hxy] ⊃ ~ Hxx} 2. (x){Hxx ⊃ (∃y)[(y ≠ x) · Hxy] 3. (x){{Sx ⋅ (∃y )[(x ≠ y) ⋅ Sxy]} ⊃ (y)[(x ≠ y) ⊃ Sxy]} 4. This sentence should be understood as a generalization about sinners who sin

against every sinner, not as a definite description: (x){[Sx ⋅ (y)(Sy ⊃ Sxy)] ⊃ Sxx} 5. (x){{Sx ⋅ (y)[(x ≠ y) ⊃ Sxy]} ⊃ Sxx} 6. (∃x){[Sx ⋅ (y)(Sy ⊃ Sxy)] ⋅ (y){[Sy ⋅ (z)(Sz ⊃ Syz)] ⊃ (x = y)}} 7. ~ (x){Sx ⊃ (∃y){[Sy ⋅ (x ≠ y)] ⋅ Syx}} 8. (x) {Sx ⊃ ~ (y) {[Sy ⋅ (x ≠ y )] ⊃ Syx}} or ~ (∃x) {Sx ⋅ (y) {[Sy ⋅ (x ≠ y)] ⊃ Syx}} 9. (x){{Sx ⋅ (y){[Sy ⋅ (x ≠ y)] ⊃ Sxy}} ⊃ (z)[{Sz ⋅ (y){[Sy ⋅ (y ≠ z)] ⊃ Szy}} ⊃ (x = z)]} 10. (∃x) {Sx ⋅ (y) {[Sy ⋅ (x ≠ y)] ⊃ ~ Syx}} ⊃ (x) [~ Sx ⊃ ~ (∃y) (Sy ⋅ Syx)] 11. Gs ⋅ (x){[Gx ⋅ (x ≠ s)] ⊃ Lsx} 12. (∃x) {(Cxo ⋅ (y) [Cyo ⊃ (y = x)]) ⋅ Rx} 13. (∃x) {(Cxo ⋅ (y) [Cyo ⊃ (y = x)]) ⋅ (x = c)} or Cco ⋅ (y) [Cyo ⊃ (y = c)] 14. (∃x) {(Cxo ⋅ (y) [Cyo ⊃ (y = x)]) ⋅ ~ (y) [(Iyo ⋅ (x ≠ y)) ⊃ Lxy]} 15. (x){Sx ⊃ (∃y)(∃z){[(Syx ⋅ Szx) ⋅ (y ≠ z)] ⋅ (w)(Swx ⊃ [(w = y) ∨ (w = z)])}}