Answers to Copi's Translation and Derivation Exercises
Polyadic Predicate Logic
Peter Suber, Philosophy Department, Earlham College

References to Irving Copi, Symbolic Logic, are to the fifth edition, Macmillan, 1979.

I include the translation exercises, but not the derivation exercises, that Copi answers himself in the back of the book.

Remember that there are many ways to prove a valid argument valid with a derivation. If my way differs from your way, yours may still be perfectly good.

I welcome corrections.

If the steps of a proof are too long for their box and wrap to a second line, then the justifications to their right may become out of step with the lines they justify. To fix this, widen your browser window with your mouse.

Table of Contents

Translations at pp. 127-28
Part I

These answers would be more helpful if I also typed in the notation formulas being translated. But that would only be helpful if I also typed in the large dictionary of predicates on p. 127 and I don't have time for that right now. Sorry.

  1. Every dog has his day.
  2. If everything is fair for someone, then everyone is fair for her.
  3. A rolling stone gathers no moss.
  4. God helps (all) those who help themselves.
  5. All things come to him who waits.
  6. There's no place like home.
  7. Those who don't lose God, don't lose anything.
  8. If you can't command yourself, then you can't command anyone.
  9. Every cloud has a silver lining.
  10. A person is judged by the company he keeps.
  11. Where there's smoke, there's fire.
  12. A jack of all trades is a master of none.
  13. People who live in glass houses shouldn't throw stones.
  14. People who like everything they say will hear something they don't like.
  15. It's an ill wind that blows nobody any good.
  16. Nothing ventured, nothing gained.
  17. Work belonging to all time is being done (at or by) every moment.
  18. God tempers all winds to the lamb somebody sheared.
  19. Everyone borrows something from someone.
  20. Nobody borrows everything from everybody.
  21. Nobody borrows everything from everybody.
  22. There is something that nobody has borrowed from anyone.
  23. All good dogs like their masters.
  24. Everyone has a parent, but some people are not parents of anyone.
  25. If people everywhere who know nothing are blissful, then people everywhere who know everything are fools.

Translations at pp. 128-29
Part II
  1. Dead men tell no tales. (x)[(Dx · Mx) (y)(Ty ~Txy)]
  2. A lawyer who pleads his own case has a fool for a client. (x)(y)[(Lx · Pxx) (Cyx · Fy)]
  3. A dead lion is more dangerous than a live dog. (x)(y)[(~Ax · Lx · Ay · Dy) Dxy]
  4. Uneasy lies the head that wears the crown. (x)(y)[(Hx · Cy · Wxy) Ux]
  5. If a boy tells only lies, none of them will be believed. (x)[[Bx · (y)(Txy Ly)] (z)[Txy (u)~Buz]]
  6. Anyone who consults a psychiatrist ought to have his head examined. (x)(y)[(Px · Sy · Cxy) Ox]
  7. No one ever learns anything unless he teaches it to himself. (x)(y)[(Px ~Lxy) Txyx]
  8. Delilah wore a ring on every finger and had a finger in every pie. (x)(y)[(Fxd · Ry) (Oyx] · (x)(y)[(Fxd · Py) Ixy]
  9. Any man who hates children and dogs cannot be all bad. (x)(y)(z)[(Mx · Cy · Dz · Hxy · Hxz) ~Bx]
  10. Anyone who accomplishes anything will be envied by everyone. (x)[Px [(y)Axy (z)(Pz Ezx)]]
  11. To catch a fish, one must have some bait. (x)(y)(z)[(Px · Fy · Bz · Cxy) Hxz]
  12. Every student does some problems, but no student does all of them. (x)(y)[(Sx · Py) Dxy] · ~(x)(y)(Sx · Py · Dxy)
  13. Any contestant who answers all the questions put to him will win any prize he chooses. (x)(y)(z)[[(Cx · Qy · Pz · Pyx) Axy] (Cxz Wxz)]
  14. Every son has a father, but not every father has a son. (x)(y)[(Px · Mx · Py · My) Pyx] · (x)(y)(Px · Mx · Py · My · ~Pxy)
  15. A person is maintaining a nuisance if he has a dog that barks at everyone who visits its owner. (x)[Px [(y)[Dy · Hxy · (z)[(Pz · Vzx) Byz]] (u)(Nu · Mxu)]]]
  16. A doctor has no scruples who treats a patient who has no ailment. (x)(y)(z)(w)[(Dx · Sy · Pz · Aw · Txz · ~Hzw) ~Hxy]
  17. A doctor who treats a person who has every ailment has a job for which no one would envy him. (x)(y)(z)(w)(u)[(Dx · Py · Txy · Az · Hyz · Jw · Pu) (Hxw · ~Euxw)]
  18. If a farmer keeps only hens, none of them will lay eggs that are worth setting. (x)(y)(z)[(Fx · Hy · Kxy · Ez) (Lyz · ~Wz)]

Translations at p. 129

For problems 19-25 (p. 129), Copi asks us to use these abbreviations:

  1. Everyone buys something from some store (or other). (x)(y)(z)[(Px · Sy) Bxzy].
  2. There is a store from which everyone buys something (or other). (x)(y)(z)[(Sx · Py) Byzx].
  3. Some people make all their purchases from a single store. (x)(y)(z)[(Px · Sz) Bxyz].
  4. No one buys everything it [sic] sells from any store. (x)(y)(z)[(Px · Sy) ~Bxzy].
  5. No one buys things from every store. (x)(y)(z)[(Px · Sy) ~Bxyz].
  6. No store has everyone for a customer. (x)(y)(z)[(Sx · Py) ~Byzx].
  7. No store makes all its sales to a single person. (x)[Sx (y)[Py (z)[(w)Bwzx · ~Byzx]]]

Translations at pp. 129-30

For problems 26-45 (pp. 129-30), Copi asks us to use these abbreviations:

  1. Nobody donates to every charity. (x)(y)(z)[(Px · Cy) ~Dxzy].
  2. Nobody donates money to every charity. (x)(y)(z)[(Px · My · Cz) ~Dxyz].
  3. Nobody donates all of his money to charity. (x)(y)(z)[(Px · My · Byx · Cz) ~Dxyz].
  4. Nobody donates all of his money to any single charity. (x)(y)(z)[(Px · My · Byx · Cz) ~Dxyz].
  5. Nobody donates all of his belongings to any single charity. (x)(y)(z)[(Px · Byx · Cz) ~Dxyz].
  6. Nobody gives all of his donations to any single charity. (x)(y)(z)[(Px · Cz) ~Dxyz].
  7. No charity receives all of its money from any single person. (x)(y)(z)[(Cx · My · Pz · Byx) ~Dzyx].
  8. No charity receives all of his [sic] money from any single person. (x)(y)(z)[(Cx · My · Pz · Byz) ~Dzyx].
  9. No charity receives all of his [sic] donations from any single person. (x)(y)(z)[(Cx · Pz) ~Dzyx].
  10. No charity receives all of his [sic] donations from any one donor. (x)(y)(z)(Cx ~Dzyx).
  11. No charity receives only money as donations. (x)(y)(z)(Cz · ~Mx · Dyxz).
  12. Somebody gives money to charity. (x)(y)(z)(Px · My · Cz · Dxyz).
  13. Somebody donates all of his money to charity. (x)(y)(z)(Px · My · Byx · Cz · Dxyz).
  14. At least one person donates all of his belongings to a single charity. (x)(y)(z)(Px · Byx · Cz · Dxyz).
  15. At least one person gives all of his donations to a single charity. (x)(y)(z)(Px · Cz · Dxyz).
  16. Some charities receive donations from everybody. (x)(y)(z)(Cx · Pz · Dzyx).
  17. Some charities receive donations from every donor. (x)(y)(z)(Cx · Dzyx).
  18. Some donations are not given to a charity. (x)(y)(z)(~Cz · Dyxz).
  19. Some donors donate to every charity. (x)(y)(z)(Cz · Dxyz).
  20. Every charity receives donations from at least one donor. (x)(y)(z)(Cx Dzyx).

Translations at p. 130
Part III
  1. Whoso sheddeth man's blood, by man shall his blood be shed. (x)[[Px · (y)[My · (z)(Bz · Bzy · Sxz)]] (u)[Mu · (w)(Bw · Bwx · Suw)]]
  2. His hand will be against every man, and every man's hand against him. (x)(y)(z)[(Px · Hy · Mz · Byx) Ayz] · [(Px · Hy · Mz · Byz) Ayx]
  3. The sun shall not smite thee by day, nor the moon by night. ~Sstd · ~Smtn
  4. A wise son maketh a glad father. (x)(y)[(Mx · Wx) (Mxy · Fyx · Gy)]
  5. He that spareth his rod hateth his son. (x)[[Mx · (y)[(Ry · Byx) Sxy]] (z)[(Mz · Pxz) Hxz]]
  6. The borrower is servant to the lender. (x)(y)(Bxy Sxy)
  7. Whose diggeth a pit shall fall therein: and he that rolleth a stone, it will return upon him. (x)(y)[(Px · Hy · Dxy) Fxy] · [(x)(y)(Px · Sy · Rxy) Byx]
  8. The fathers have eaten sour grapes, and the children's teeth are set on edge. (x)(y)(z)(w)[(Fxy · Gz · Sz · Exz · Tw) (Bwy · Ew)]
  9. The foxes have holes, and the birds of the air have nests; but the Son of man hath not where to lay his head. (x)(y)[(Fx · Hy) Hxy] · (x)(y)[(Bx · Ny) Hxy] · (x)(y)(z)[(Sx · Cy · Byx) ~Pzxy]
  10. the good that I would, I do not; but the evil which I would not, that I do. (x)[(Gx · Wix) ~Dix] · (y)[(Ey · ~Wiy) Diy]

Derivations at pp. 132-33


Part I, exercise 2 (p. 132)
1. (x)[(y)Byz (z)Bxz]
2. Byz
3. (x)[(y)Byx Bxy
4. (y)Byz Bzy
5. (y)Byz
6. Bzy
7. Byz Bzy
8. (z)(Byz Bzy)
9. (y)(z)(Byz Bzy)
/ (y)(z)(Byz Bzy)
ass. for CP
1 UI
3 UI
2 EG
4,5 MP
2-6 CP
7 UG
9 UG
Once we discover that we must instantiate 'x' to 'z', and 'z' to 'y', the trick here was to instantiate the 'z' quantifier before the 'x' quantifier. Otherwise we'd get something like (z)Bzz and have to instantiate it to something like Bzy, which would violate the convention.

Part I, exercise 3 (p. 133)
1. (x)(Cax Dxb)
2. (x)Dxb (y)Dby
3. (x)Cax
4. Cax
5. Cax Dxb
6. Dxb
7. (x)Dxb
8. (y)Dby
9. (y)Dby
10. (x)Cax (y)Dby

/ (x)Cax (y)Dby
ass. for CP
3 EI, ass.
1 UI
4,5 MP
6 EG
2,7 MP
3,4-8 EI
3-9 CP


Part I, exercise 4 (p. 133)
1. (x)[Ex (y)(Fy Gxy)]
2. (x)[Ex · (y)~Gxy]
3. Ex · (y)~Gxy
4. (y)~Gxy
5. ~Gxy
6. Ex (y)(Fy Gxy)
7. Ex
8. (y)(Fy Gxy)
9. Fy Gxy
10. ~Fy
11. (x)~Fx
12. (x)~Fx
13. (x)~Fx

/ (x)~Fx
2 EI, ass.
3 simp
4 EI, ass.
1 UI
3 simp
6,7 MP
8 UI
5,9 MT
10 EG
4,5-11 EI
2,3-12 EI


Part I, exercise 6 (p. 133)
1. (x)[Kx [(y)Lxy (z)Lzx]]
2. (x)[(z)Lzx Lxx]
3. ~(x)Lxx
4. Kx
5. (x)~Lxx
6. ~Lxx
7. (z)Lzx Lxx
8. ~(z)Lzx
9. Kx [(y)Lxy (z)Lzx]
10. (y)Lxy (z)Lzx
11. ~(y)Lxy
12. (y)~Lxy
13. Kx (y)~Lxy
14. (x)(Kx (y)~Lxy)


/ (x)(Kx (y)~Lxy)
ass. for CP
3 QN
5 UI
2 UI
6,7 MT
1 UI
4,9 MP
8,10 MT
11 QN
4-12 CP
13 UG


Part I, exercise 7 (p. 133)
1. (x)[Mx (y)(Ny Oxy)]
2. (x)[Px (y)(Oxy Qy)]
3. (x)(Mx · Px)
4. Mx · Px
5. Mx
6. Px
7. Mx (y)(Ny Oxy)
8. Px (y)(Oxy Qy)
9. (y)(Ny Oxy)
10. (y)(Oxy Qy)
11. Ny Oxy
12. Oxy Qy
13. Ny Qy
14. (y)(Ny Qy)
15. (y)(Ny Qy)
16. (x)(Mx · Px) (y)(Ny Qy)

/ (x)(Mx · Px) (y)(Ny Qy)
ass. for CP
3 EI, ass.
4 simp
4 simp
1 UI
2 UI
5,7 MP
6,8 MP
9 UI
10 UI
11,12 HS
13 UG
3,4-14 EI
3-15 CP
The use of UG in line 14 is permissible because we are generalizing on 'y' which is not free in either of our assumptions.

Part I, exercise 8 (p. 133)
1. (x)[(Rx · ~Sx) (y)(Txy · Uy)]
2. (x)[Vx · Rx · (y)(Txy Vy)]
3. (x)(Vx ~Sx)
4. Vx · Rx · (y)(Txy Vy)
5. (Rx · ~Sx) (y)(Txy · Uy)
6. Vx ~Sx
7. Vx
8. Rx
9. (y)(Txy Vy)
10. ~Sx
11. Rx · ~Sx
12. (y)(Txy · Uy)
13. Txy · Uy
14. Txy Vy
15. Txy
16. Vy
17. Uy
18. Vy · Uy
19. (x)(Vx · Ux)
20. (x)(Vx · Ux)
21. (x)(Vx · Ux)


/ (x)(Vx · Ux)
2 EI, ass.
1 UI
3 UI
4 simp
4 simp
4 simp
6,7 MP
8,10 conj
12,5 MP
12 EI, ass.
9 UI
13 simp
14,15 MP
13 simp
16,17 conj
18 EG
12,13-19 EI
2,4-20 EI
The trick here is to resist the temptation to use the Vx in line 7 in order to build up to the conclusion; rather, use the Vy in line 16. The use of EI is permissible in line 13 because we are instantiating to 'y', which is not yet free in the proof at that point.

Part I, exercise 9 (p. 133)
1. (x)(Wx Xx)
2. (x)[(Yx · Xx) Zx]
3. (x)(y)(Yy · Ayx)
4. (x)(y)[(Ayx · Zy) Zx]
5. (y)(Ayx Wy)
6. (x)(Yy · Ayx)
7. Wy Xy
8. (Yy · Xy) Zy
9. (y)[(Ayx · Zy) Zx]
10. (Ayx · Zy) Zx
11. Ayx Wy
12. Yy · Ayx
13. Yy
14. Ayx
15. Wy
16. Xy
17. Yy · Xy
18. Zy
19. Ayx · Zy
20. Zx
21. Zx
22. (y)(Ayx Wy) Zx]
23. (x)[(y)(Ayx Wy) Zx]



/ (x)[(y)(Ayx Wy) Zx]
ass. for CP
3 EI, ass.
1 UI
2 UI
4 UI
9 UI
5 UI
6 UI
12 simp
12 simp
11,14 MP
7,15 MP
13,16 conj
8,17 MP
14,18 conj
10,19 MP
3,6-20 EI
5-21 CP
22 UG
The trick here is instantiate 'x' in lines 1 and 2 to one letter (here 'y' in lines 7 and 8) and later in lines 4, 5, 6, and 9, to instantiate 'x' to a different letter (here 'x' in lines 9, 10, 11, and 12), even though it would have been legal to instantiate them all to the same letter. This is a case when we have to violate the rule of thumb to instantiate to the same letter whenever we can. Incidentally, it's permissible to discharge our EI assumption at line 20 because, although line 20 has a free 'x' in it, there is no free 'x' in the EI assumption itself (line 6).

Part I, exercise 10 (p. 133)
1. (x)[[Bx · (y)[Cy · Dyx · (z)(Ez · Fxz)]] (w)Gxwx]
2. (x)(y)(Hxy Dyx)
3. (x)(y)(Fxy Fyx)
4. (x)(Ix Ex)


5. Bx
6. (y)(Cy · Hxy) · (z)(Iz · Fxz)
7. (y)(Cy · Hxy)
8. (z)(Iz · Fxz)
9. Cy · Hxy
10. Iz · Fxz
11. Bx · (y)[Cy · Dyx · (z)(Ez · Fxz)] (w)Gxwx
12. (y)[Cy · Dyx · (z)(Ez · Fxz)] (w)Gxwx
13. (y)(Hxy Dyx)
14. Hxy Dyx
15. (y)(Fxy Fyx)
16. Fxz Fzx
17. Iz Ez
18. Cy
19. Hxy
20. Dyx
21. Iz
22. Ez
23. Fxz
24. Ez · Fxz
25. (z)(Ez · Fxz)
26. Cy · Dyx
27. Cy · Dyx · (z)(Ez · Fxz)
28. (y)[Cy · Dyx · (z)(Ez · Fxz)]
29. (w)Gxwx
30. (u)(w)Gxwu
31. (u)(w)Gxwu
32. (u)(w)Gxwu
33. [(y)(Cy · Hxy) · (z)(Iz · Fxz)] (u)(w)Gxwu
34. Bx [[(y)(Cy · Hxy) · (z)(Iz · Fxz)] (u)(w)Gxwu]
35. (x)[Bx [[(y)(Cy · Hxy) · (z)(Iz · Fxz)] (u)(w)Gxwu]]



/ (x)[Bx [[(y)(Cy · Hxy) ·
(z)(Iz · Fxz)]
(u)(w)Gxwu]]
ass. CP
ass. CP
6 simp
6 simp
7 EI, ass.
8 EI, ass.
1 UI
11 simp
2 UI
13 UI
3 UI
15 UI
4 UI
9 simp
9 simp
14,19 MP
10 simp
17,21 MP
10 simp
22,23 conj
24 EG
18,20 conj
26,25 conj
27 EG
28,12 MP
29 EG
8,10-30 EI
7,9-31 EI
6-32 CP
5-33 CP
34 UG
The hardest trick here was to see that we could perform step 30: generalize one of the x's in step 29 without generalizing the other one. The convention permits this, although it is only rarely useful to do.

Translations + derivations at pp. 133-34
Part II

Part II, exercise 1 (p. 133)
1. (x)(Sxi Vxj)
2. (x)(Vax Fxh)
3. (x)(Fxk ~Fjx)
4. Fhk ~Fjh
5. Vaj Fjh
6. Fjh ~Fhk
7. Vaj ~Fhk
8. Sai Vaj
9. Sai ~Fhk
10. Fhk ~Sai


/ Fhk ~Sai
2 UI
3 UI
4 trans
5,6 HS
1 UI
7,8 HS
9 trans
Translating the second premise is difficult. Paraphrase: Anderson will vote for x only if x is a friend of Harris. If Anderson will vote for x, then x is a friend of Harris. Translating the third premise is also difficult. I'm assuming that in Fxy, x is the active befriender and y is the passive befriended. Paraphrase: no active befriender of Kelly is passively befriended by Jones. However, the English also supports, though not quite as well, this reading: no active befriender of Kelly is an active befriender of Jones.

Part II, exercise 2 (p. 133)
1. (x)(Cx Fx)
2. Cx · Dyx
3. Cx
4. Dyx
5. Cx Fx
6. Fx
7. Fx · Dyx
8. (Cx · Dyx) (Fx · Dyx)
9. (y)[(Cx · Dyx) (Fx · Dyx)]
10. (x)(y)[(Cx · Dyx) (Fx · Dyx)]
/ (x)(y)[(Cx · Dyx) (Fx · Dyx)]
ass. for CP
2 simp
2 simp
1 UI
3,5 MP
6,4 conj
2-7 CP
8 UG
9 UG


Part II, exercise 3 (p. 133)
1. (x)(Fxa Fxb)
2. Px · Kxy · Fya
3. Px
4. Kxy
5. Fya
6. Fya Fyb
7. Fyb
8. Kxy · Fyb
9. (Px · Kxy · Fya) (Kxy Fyb)
10. (y)[(Px · Kxy · Fya) (Kxy Fyb)]
11. (x)(y)[(Px · Kxy · Fya) (Kxy Fyb)]
/ (x)(y)[(Px · Kxy · Fya) (Kxy Fyb)]
ass. for CP
2 simp
2 simp
2 simp
1 UI
5,6 MP
4,7 conj
2-9 CP
9 UG
10 UG


Part II, exercise 5 (p. 133)
1. (x)(y)[(Cx · Ey) Wxy]
2. (x)(y)(Cx · ~Cy · ~Wxy)
3. (y)(Cx · ~Cy · ~Wxy)
4. Cx · ~Cy · ~Wxy
5. (y)[(Cx · Ey) Wxy]
6. (Cx · Ey) Wxy
7 Cx
8. ~Cy
9. ~Wxy
10. ~(Cx · Ey)
11. ~Cx ~Ey
12. ~Ey
13. ~Cy · ~Ey
14. (x)(~Cx · ~Ex)
15. (x)(~Cx · ~Ex)
16. (x)(~Cx · ~Ex)

/ (x)(~Cx · ~Ex)
2 EI, ass
3 EI, ass
1 UI
5 UI
4 simp
4 simp
4 simp
6,9 MT
10 DeM
7,11 DS
8,12 conj
13 EG
3,4-14 EI
2,3-15 EI
(1) The second premise, paraphrase: At least one member of the Country Club is not wealthier than at least one non-member. At least one non-member is wealthier than at least one member. (2) The conclusion, paraphrase: At least one person belongs to neither the Country Club nor the Elks Lodge.

Part II, exercise 7 (pp. 133-34)
1. (x)(Dx Mx)
2. (x)(y)[(Px · My · Wxy) Gx]
3. (x)(y)(Px · Ox · Ny · Dy · Wxy)
4. (y)(Px · Ox · Ny · Dy · Wxy)
5. Px · Ox · Ny · Dy · Wxy
6. Dy My
7. Dy
8. My
9. Px
10. Wxy
11. Px · My · Wxy
12. (y)[(Px · My · Wxy) Gx]
13. (Px · My · Wxy) Gx
14. Gx
15. Ox
16. Px · Ox · Gx
17. (x)(Px · Ox · Gx)
18. (x)(Px · Ox · Gx)
19. (x)(Px · Ox · Gx)


/ (x)(Px · Ox · Gx)
3 EI, ass.
4 EI, ass.
1 UI
4 simp
6,7 MP
5 simp
5 simp
9,8,10 conj conj
2 UI
12 UI
11,13 MP
5 simp
9,15,14 conj conj
16 EG
4,5-17 EI
3,4-18 EI

I haven't had time to finish this set of exercises. I hope to do so soon. . . .

Part II, exercise 8 (p. 134)


1. (x)(y)[(Px · Sy · Lxy) Lyx]
2. (x)(y)[(Sx · Py) Lxy]
/ (x)(y)[(Px · Sy) Lyx]


Part II, exercise 10 (p. 134)
1. (x)(y)[(Px · Dxyf) (y)Dxyf]
/ (x)(Pj · ~Djxf) (x)~Djxf


Part II, exercise 11 (p. 134)
1. (x)(y)(z)[(Bx · Cy · Axy · Pz · Lz) Rzx]
2. (x)(y)[(Px · Rxy) Txy]
or: (x)(y)[(Px · Rxy) Txy]
3. (x)(y)(z)[(Cx · By · Pz · Wzy · Fzx) Axy]



/ (x)(y)(z)(w)[(Px · Cy · Fxy · Bz · Pw · Lw) (Wxz Twz)]


Part II, exercise 12 (p. 134)

1. (x)(y)(z)[(Wx · Sy · Txy · Pz) Uxz]
2. (x)(y)(Wx · Rx · Ay · Cyx)
3. (x)[(Wx · Rx · Sy · Iy) Txy]

/ (x)(y)[(Px · Axy) ~Uxy](x)(y)(z)(Ax · Cxy · ~Azy)



This file is an electronic hand-out for the course, Symbolic Logic.

Most of the logic symbols in this file are GIFs. See my Notes on Logic Notation on the Web.

[Blue
Ribbon] Peter Suber, Department of Philosophy, Earlham College, Richmond, Indiana, 47374, U.S.A.
peters@earlham.edu. Copyright © 1997, Peter Suber.