Translating Sentences

Translations at pp. 69-70

1. Snakes are reptiles. Answer
   • ∀X (s(X) ⇒ r(X))
2. Snakes are not all poisonous. Answer
   • ∃X (s(X) & ~p(X))
   • This says there is at least one snake that is not poisonous; this commits us to the existence of snakes, which the author seems to want.
   • Don't write ∃X ~(s(X) & p(X)), or ~∀X (s(X) ⇒ p(X)), which say instead that there is at least one thing that is not a poisonous snake --a claim satisfied by my chalk.
3. Children are present. Answer
   • ∃X (c(X) & p(X))
   • My translation says that at least one child is present, which seems more accurate than ∀X (c(X) ⇒ p(X)), all children in the universe (of discourse) are present.
4. Executives all have secretaries. Answer
   • ∀X (e(X) ⇒ s(X))
5. Only executives have secretaries. Answer
   • ∀X (s(X) ⇒ e(X))
   • Compare the answers to 4 and 5; note how the 'all' and 'only' propositions are the converse of one another.
6. Only property owners may vote in special municipal elections. Answer
   • ∀X (v(X) ⇒ p(X))
   • If you wrote ∀X (p(X) ⇒ v(X)), then you'd be saying that all (not only) property owners may vote.
7. Employees may use only the service elevator. Answer
   • ∀X (u(X) ⇒ s(X))
   • Paraphrase: All elevators which employees may use are service elevators.
8. Only employees may use the service elevator. Answer
   • ∀X (u(X) ⇒ e(X))
   • Paraphrase: All those who may use service elevators are employees.
   • Note that in problem #7, u(X) referred to elevators, and in problem #8, u(X) refers to employees.
9. All that glitters is not gold. Answer
   • ∀X (g(X) ⇒ ~a(X))
   • My translation, like the English original, says quite literally that all that glitters is not gold. It implies that gold, if it glitters, is not gold, which is false.
   • The speaker of this proverb undoubtedly meant that we should not be misled by glitter to think that we have found gold. This truth requires a 'not all' rather than an 'all not', and should be translated ~∀X (g(X) ⇒ a(X)) or ∃X (g(X) & ~a(X)).
   • There are two natural but non-equivalent ways to translate 'all not' sentences, such as 'All that glitters is not gold': ∀X ~(g(X) ⇒ a(X)) and ∀X (g(X) ⇒ ~a(X)). The former asserts the absurdity that everything whatsoever glitters and everything whatsoever is not gold. The latter is the true 'all not' sentence: for all things, if they glitter, then they are not gold. Remember that "All A's are not B's" means "No A's are B's".
10. None but the brave deserve the fair. Answer
    • ∀X (d(X) ⇒ b(X))
    • This means: only the brave deserve the fair, which is the converse of all the brave deserve the fair. It is equivalent to saying: All those who deserve the fair are brave.
11. Not every visitor stayed for dinner. Answer
    • ∃X (v(X) & ~s(X))
    • My translation says there is a visitor who did not stay; this seems more accurate to me than ∃X ~(v(X) & s(X)), or ~∀X (v(X) ⇒ s(X)): there is something that is not a visitor who stayed, for example, my chalk.
12. Not any visitor stayed for dinner. Answer
    • ∀X (v(X) ⇒ ~s(X)), or ~∃X (v(X) & s(X))
    • 'Not any' = none = all did not = false that at least one did.
13. Nothing in the house escaped destruction. Answer
    • ∀X ~(h(X) ⇒ e(X)) and ∀X (h(X) ⇒ ~e(X)) are both natural; but they are non-equivalent (see note to exercise 9). The former says that being in the house didn't necessarily mean destruction, clearly not the speaker's meaning. The latter says that everying in the house was destroyed.
14. Some students are both intelligent and hard workers. Answer
    • ∃X (s(X) & i(X) & h(X))
    • There is at least one thing possessing these three properties: being a student, being intelligent, being a hard worker. (This notation is a natural for poetry!)
15. No coat is waterproof unless it has been specially treated. Answer
    • ∀X (c(X) ⇒ (~w(X) | s(X)))
    • If something is a coat, then it's either not waterproof or specially treated. This is an accurate translation; you'll have to get used to turning sentences inside out like this on occasion.
    • Here I translated "unless" as inclusive disjunction. If you want to use exclusive disjunction, then say this: ∀X (c(X) ⇒ (~w(X) <~> s(X)))
16. Some medicines are dangerous only if taken in excessive amounts. Answer
    • ∃X (m(X) & (d(X) ⇒ e(X)))
    • This way of translating it commits us absolutely to the existence of medicines, but only conditionally to their dangerousness, which is what the speaker seems to want.
    • We avoid ∃X ((m(X) & d(X)) ⇒ e(X)) because it is an existentially quantified conditional. It is equivalent to ∃X (~(m(X) & d(X)) | e(X)) which asserts that there is something that is either not a dangerous medicine (like my chalk) or that is taken in excessive amounts (like logic courses).
17. All fruits and vegetables are wholesome and delicious. Answer
    • ∀X ((f(X) | v(X)) ⇒ (w(X) & d(X)))
    • Note how the "and" between fruits and vegetables is translated as a disjunction.
18. Everything enjoyable is either immoral, illegal, or fattening. Answer
    • ∀X (e(X) ⇒ (~m(X) | ~l(X) | f(X)))
19. A professor is a good lecturer if and only if he is both well-informed and entertaining. Answer
    • ∀X (p(X) ⇒ (g(X) ⇔ (w(X) & e(X))))
    • My translation is not equivalent to this attractive possibility: ∀X ((p(X) & g(X)) ⇔ (w(X) & e(X))). Mine says that all professors are good iff they are well-informed and entertaining. The equivalence is only asserted to hold for professors. The latter says that something is a good professor iff it is well-informed and entertaining; the equivalence is asserted to hold in general, implying that the puerile weather person on the local news is a 'good professor'.
20. Only policemen and firemen are both indispensable and underpaid. Answer
    • ∀X ((i(X) & u(X)) ⇒ (p(X) | f(X)))
    • Paraphrase: All who are indispensable and underpaid are either policemen or firemen. Note how the "and" between policemen and firemen is translated as a disjunction.
21. Not every actor is talented who is famous. Answer
    • ∃X (a(X) & f(X) & ~t(X))
    • At least one actor who is famous is not talented.
22. Any girl is healthy if she is well nourished and exercises regularly. Answer
    • ∀X (g(X) ⇒ ((w(X) & e(X)) ⇒ h(X))
    • This is equivalent to ∀X ((g(X) & w(X) & e(X)) ⇒ h(X)) by exportation.
23. It is not true that every watch will keep good time if and only if it is wound regularly and not abused. Answer
    • ~∀X (w(X) ⇒ ((r(X) & ~a(X)) ⇔ k(X))) or ∃X ~(w(X) ⇒ ((r(X) & ~a(X)) ⇔ k(X)))
    • See comment to exercise 19.
    • Note that this is not an existentially quantified conditional, but an existentially quantified negation of a conditional.
24. Not every person who talks a great deal has a great deal to say. Answer
    • ∃X (p(X) & t(X) & ~h(X))
    • At least one person who talks a great deal does not have a great deal to say.
25. No automobile that is over ten years old will be repaired if it is severely damaged. Answer
    • ∀X ((a(X) & o(X)) ⇒ (d(X) ⇒ ~r(X)))
    • This is equivalent to ∀X ((a(X) & o(X) & d(X)) ⇒ ~r(X)) by exportation.
    • It is not equivalent to ∀X ~((a(X) & o(X) & d(X)) ⇒ r(X)).

    Translations at p. 70

26. Some horses are gentle and have been well trained. Answer
    • ∃X (h(X) & g(X) & t(X))
27. Some horses are gentle only if they have been well trained. Answer
    • ∃X (h(X) & (g(X) ⇒ t(X)))
    • Don't forget that "p ⇒ q" translates "p only if q".
28. Some horses are gentle if they have been well trained. Answer
    • ∃X (h(X) & (t(X) ⇒ g(X)))
29. Any horse is gentle that has been well trained. Answer
    • ∀X (h(X) ⇒ (t(X) ⇒ g(X)))
    • Paraphrase: All well trained horses are gentle.
    • This is equivalent to ∀X ((h(X) & t(X)) ⇒ g(X)) by exportation
30. Any horse that is gentle has been well trained. Answer
    • ∀X (h(X) ⇒ (g(X) ⇒ t(X)))
    • Paraphrase: All gentle horses are well trained.
    • This is equivalent to ∀X ((h(X) & g(X)) ⇒ t(X))
31. No horse is gentle unless it has been well trained. Answer
    • ∀X (h(X) ⇒ (g(X) ⇒ t(X)))
    • Paraphrase: All horses are ungentle unless well trained. All horses are ungentle or (else) well trained.
    • Note that this is the same formula used in exercise 30. The trick is to see that the statement in 31 asserts that being well trained is the necessary condition of being gentle. Recall that necessary conditions are translated as the consequents of ordinary conditionals; that which they condition becomes the antecedent. You may find this translation more intuitive if you transform it into one or more of its equivalents, e. g. ∀X (h(X) ⇒ (~t(X) ⇒ ~g(X))) or ∀X (h(X) ⇒ (t(X) | ~g(X))) or ∀X ((h(X) & ~t(X)) ⇒ ~g(X)).
32. Any horse is gentle if it has been well trained. Answer
    • ∀X (h(X) ⇒ (t(X) ⇒ g(X)))
    • Or by exportation: ∀X ((h(X) & t(X)) ⇒ g(X)).
33. Any horse has been well trained if it is gentle. Answer
    • ∀X (h(X) ⇒ (g(X) ⇒ t(X)))
    • Or by exportation: ∀X ((h(X) & g(X)) ⇒ t(X)).
34. Any horse is gentle if an only if it has been well trained. Answer
    • ∀X (h(X) ⇒ (g(X) ⇔ t(X)))
35. Gentle horses have all been well trained. Answer
    • ∀X (h(X) ⇒ (g(X) ⇒ t(X)))
    • Or by exportation: ∀X ((h(X) & g(X)) ⇒ t(X)).
    • Note that this is the same formula used in exercise 33.
36. Only well trained horses are gentle. Answer
    • ∀X (h(X) ⇒ (g(X) ⇒ t(X)))
    • Or by exportation: ∀X ((h(X) & g(X)) ⇒ t(X)).
    • Note that this is the same formula used in exercises 33 and 35.
37. Only gentle horses have been well trained. Answer
    • ∀X (h(X) ⇒ (t(X) ⇒ g(X)))
    • Or by exportation: ∀X ((h(X) & t(X)) ⇒ g(X)).
    • Note that this is the same formula used in exercise 32.
38. Only horses are gentle if they have been well trained. Answer
    • I don't know what this means. If it means "horses are the only things that can be made gentle by training," then translate it thus: ∀X ((t(X) ⇒ g(X)) ⇒ h(X)) (if something became gentle from training, then it's a horse).
39. Some horses are gentle even though they have not been well trained. Answer
    • ∃X (h(X) & g(X) & ~t(X))
40. If something is a well trained horse, then it must be gentle. Answer
    • ∀X ((h(X) & t(X)) ⇒ g(X))
    • Despite the 'some' in 'something', we need the universal quantifier here. The speaker clearly means that every well trained horse is gentle, not that at least one well trained horse is gentle.
    • Note that this is the same formula used in exercises 32 and 37.
41. Some horses that are well trained are not gentle. Answer
    • ∃X (h(X) & t(X) & ~g(X))
42. Some horses are neither gentle nor well trained. Answer
    • ∃X (h(X) & ~g(X) & ~t(X))
43. No horse that is well trained fails to be gentle. Answer
    • ∀X (h(X) ⇒ (t(X) ⇒ g(X)))
    • Paraphrase: All well trained horses are gentle.
    • Or by exportation: ∀X ((h(X) & t(X)) ⇒ g(X)).
    • Note that this is the same formula used in exercises 32, 37, and 40.
44. A horse is gentle only if it has been well trained. Answer
    • ∀X (h(X) ⇒ (g(X) ⇒ t(X)))
    • On 'only if', see comment to exercise 27.
    • Note that the indefinite article takes the universal quantifier here.
    • Or by exportation: ∀X ((h(X) & g(X)) ⇒ t(X)).
    • Note that this is the same formula used in exercises 33, 35, and 36.
45. If anything is a gentle horse, then it has been well trained. Answer
    • ∀X ((h(X) & g(X)) ⇒ t(X))
    • Note that this is the same formula used in exercises 33, 35, 36, and 44.
46. If any horse is well trained, then it is gentle. Answer
    • ∀X ((h(X) & t(X)) ⇒ g(X))
    • Note that this is the same formula used in exercises 32, 37, 40, and 43.

    Translations at p. 71

47. Blessed is he that considereth the poor. Answer
    • ∀X (c(X) ⇒ b(X))
    • Paraphrase: All who considereth the poor are blessed.
    • Note how the singular 'he' triggers the universal quantifier; the speaker clearly means "blessed is whoever that considereth. . . ". This is true in many of the following exercises as well.
    • Copi assumes that the universe of discourse is limited to persons, otherwise he'd have to indicate this limitation in the formula: ∀X (p(X) ⇒ (c(X) ⇒ b(X))). This is true in many of the following exercises as well.
48. He that hath knowledge spareth his words. Answer
    • ∀X (h(X) ⇒ s(X))
    • Paraphrase: All who hath, spareth.
49. Whoso findeth a wife findeth a good thing. Answer
    • ∀X (w(X) ⇒ g(X))
    • Paraphrase: All who find wives, find good things.
50. He that maketh haste to be rich shall not be innocent. Answer
    • ∀X (r(X) ⇒ ~i(X))
    • Paraphrase: None who make haste to be rich are innocent.
51. They shall sit every man under his vine and under his fig-tree. Answer
    • ∀X (m(X) ⇒ (v(X) & f(X)))
    • Paraphrase: If one is a man, they shall sit one under one's vine and they shall sit one under one's fig-tree.
    • It doesn't matter who 'they' are; that can be built into the V and F predicates. v(X) = 'they' shall sit x under his/her vine; f(X) = 'they' shall sit x under his/her fig-tree. If it looks like 'they' are the subjects of these sentences, whereas x ought to represent the subject, then we restate: v(X) = x shall be sat under his/her vine by them; f(X) = x shall be sat under his/her fig-tree by them.
52. He that increaseth knowledge increaseth sorrow. Answer
    • ∀X (k(X) ⇒ s(X))
    • Paraphrase: All who increase knowledge, increase sorrow.
53. Nothing is secret which shall not be made manifest. Answer
    • ∀X (s(X) ⇒ m(X))
    • Paraphrase: All that is secret shall be made manifest.
54. Whom The Lord loveth He chasteneth. Answer
    • ∀X (l(X) ⇒ c(X))
    • Paraphrase: All who are loved by the lord are chastened by the lord.
55. If a man desire the office of a bishop, he desireth a good work. Answer
    • ∀X (o(X) ⇒ w(X))
    • Paraphrase: All who desire the office of a bishop desire a good work.
56. He that hateth dissembleth with his lips, and layeth up deceit within him. Answer
    • ∀X (h(X) ⇒ (d(X) & l(X)))
    • Paraphrase: All who hate dissemble and lay up deceit.

    Translations at pp. 88-89

57. If anything is damaged, someone will be blamed. Answer
    • ∀X (d(X) ⇒ ∃Y (p(Y) & b(Y)))
    • These variations seem acceptable but are not: ∃X (d(X) ⇒ ∃Y (p(Y) & b(Y))), and ∃X ∃Y (d(X) ⇒ (p(Y) & b(Y))). The reason is that they are existentially quantified conditionals. (To see why existentially quantified conditionals are usually undesirable, see comments to exercise 2, below.) The same is true for analogous variations on the exercises below.
    • However, if we take the first of these unacceptable translations and limit the scope of the first quantifier, then we no longer have an existentially quantified conditional: ∃X d(X) ⇒ ∃Y (p(Y) & b(Y)). Is this acceptable now? This says: if at least one thing is damaged (but not necessarily for all damaged things), some one will be blamed. This is close but not as close as my translation above, which says: for all things that are damaged (if any, but with no exceptions), some one will be blamed.
58. If anything is damaged, the tenant will be charged for it. Answer
    • ∀X (d(X) ⇒ c(X))
    • Paraphrase: All things that are damaged are charged to the tenant.
    • Note that this proposition is not multiply general; it has only one quantifier.
    • Copi wanted an existential quantifier in exercise 1; by analogy one might want one here. But like most existentially quantified conditionals, ∃X (d(X) ⇒ c(X)) does not mean what we want it to. It is equivalent to ∃X (~d(X) | c(X)), which says that there is something that is either not damaged or charged to the tenant. It commits us to the existence of that something; the English does not. Moreover, it is satisfied (made true) by my chalk, since it is undamaged; and this is certainly not intended by the speaker.
    • But ∃X (d(X) & c(X)) is clearly worse, for it asserts that there is something that is damaged that has been charged to the tenant. Hence I prefer the universal quantifier here.
59. If nothing is damaged, nobody will be blamed. Answer
    • ∀X ~d(X) ⇒ ∀Y (p(Y) ⇒ ~b(Y))
    • This is equivalent to ~∃X d(X) ⇒ ∀Y (p(Y) ⇒ ~b(Y)).
60. If something is damaged, but nobody is blamed, the tenant will not be charged for it. Answer
    • ∀X ((d(X) & ∀Y (p(Y) ⇒ ~b(Y))) ⇒ ~c(X))
    • See comment to exercise 2, above. Here the "if something" initially pushes us to use an existentially quantified conditional; but since such expressions don't mean what we want them to, we retreat to a universal quantifier. Hence the paraphrase underlying our translation is: All things that are damaged when nobody is blamed will not be charged to the tenant. Or: Nothing which is damaged when nobody is blamed will be charged to the tenant.
61. If any bananas are yellow, they are ripe. Answer
    • ∀X ((b(X) & y(X)) ⇒ r(X))
    • Paraphrase: All yellow bananas are ripe.
    • See comments to exercises 2 and 4, above.
62. If any bananas are yellow, then some bananas are ripe. Answer
    • ∃X (b(X) & y(X)) ⇒ ∃Y (b(Y) & r(Y))
    • Note this is not an existentially quantified conditional. It is an unquantified conditional with an existentially quantified antecedent and consequent.
63. If any bananas are yellow, then if all yellow bananas are ripe, they are ripe. Answer
    • ∀X ((b(X) & y(X)) ⇒ (∀Y ((b(Y) & y(Y)) ⇒ r(Y)) ⇒ r(X)))
    • Paraphrase: All yellow bananas if all yellow bananas are ripe are ripe.
    • See comments to exercises 2 and 4, above. Making the x quantifier existential is tempting, but then we'd have an existentially quantified conditional.
64. If all ripe bananas are yellow, some yellow things are ripe. Answer
    • ∀X ((b(X) & r(X)) ⇒ y(X)) ⇒ ∃Y (y(Y) & r(Y))
65. If all officers present are either captains or majors, then either some captains are present or some majors are present. Answer
    • ∀X ((o(X) & p(X)) ⇒ (∀Y (m(Y) ⇒ ~p(Y)) | m(X)))
66. If any officer is present, then either no majors are present or he is a major. Answer
    • ∀X ((o(X) & p(X)) ⇒ (c(X) | m(X))) ⇒ (∃Y (c(Y) & p(Y)) | ∃Z (m(Z) & p(Z)))
    • Paraphrase: For all present officers, either none are majors or that one is a major.
    • See comments to exercises 2 and 4, above. The x quantifier cannot be existential, for then we'd have an existentially quantified conditional.
67. If some officers are present, then if all officers present are captains, then some captains are present. Answer
    • ∃X (o(X) & p(X)) ⇒ (∀Y ((o(Y) & p(Y)) ⇒ c(Y)) ⇒ ∃Z (c(Z) & p(Z)))
    • Note that this is not an existentially quantified conditional; it is an unquantified conditional with an existentially quantified antecedent.
    • If we wanted to say "c(X)" at the end of this formula, instead of introducing the new quantifier on "z", then we'd have to extend the scope of the first ∃X quantifier to the end of the formula. That would turn the whole thing into an existentially quantified conditional, which is a good reason not to do it.
    • If we make the x quantifier universal, then we could put the rest of the formula within its scope, drop the z quantifier, and change c(Z) and p(Z) at the end to c(X) and p(X). The only problem with this approach is a slight clash with the English meaning. It wouldn't say that all officers were present, but it would say that, for all officers that are present, then (if all officers present are captains) they are captains.
68. If some officers are present, then if all officers present are captains, then they are captains. Answer
    • ∀X ((o(X) & p(X)) ⇒ (∀Y ((o(Y) & p(Y)) ⇒ c(Y)) ⇒ c(X)))
    • Paraphrase: All present officers if all present officers are captains are captains.
    • The syntax of this sentence, hence its translation, is identical to that in exercise 7.
    • Here we sorely want to use an existential quantifier: ∃X ((o(X) & p(X)) ⇒ (∀Y ((o(Y) & p(Y)) ⇒ c(Y)) ⇒ c(X))). The trouble with this, however, is that creates an existentially quantified conditional. This is not one of the rare cases where we really mean what an existentially quantified conditional would say, however. It would say (after a few transformations that preserve equivalence) that there is something that is either not a present officer or that has the property of being a captain if all present officers are captains. But this sentence is satisfied by my chalk, since my chalk is not a present officer.
    • Another tempting solution would be to retain the existential quantifier but let its scope stop after the antecedent, and not envelope the whole conditional: ∃X (o(X) & p(X)) ⇒ (∀Y ((o(Y) & p(Y)) ⇒ c(Y)) ⇒ c(X))). The problem with this is that the "x" in the final c(X) is free; hence we will have produced a propositional function (lacking a truth-value), while the English sentence was a proposition (bearing a truth-value).
    • Notice that both my translation and the English sentence are non-committal on the existence of present officers. This supports the decision to use a universal quantifier.
    • Also notice how the particular-seeming "they" does not compel an existential quantifier. This is true even of the "he" in exercise 20, below. Recall this usage in other, more familiar English sentences. "The horse is a noble animal; she swats flies with her tail. " This is clearly a statement about all horses that would take a universal quantifier.
69. If all survivors are fortunate and only women were survivors, then if there are any survivors, then some women are fortunate. Answer
    • ∀X ((s(X) ⇒ f(X)) & (s(X) ⇒ w(X))) ⇒ (∃Y (s(Y)) ⇒ ∃Z (w(Z) & f(Z)))
    • We could just as legitimately use, say, ∀Z (s(Z) ⇒ w(Z)) in place of s(X) ⇒ w(X).
70. If any survivors are women, then if all women are fortunate, they are fortunate. Answer
    • ∀X ((s(X) & w(X)) ⇒ (∀Y (w(Y) ⇒ f(Y)) ⇒ f(X)))
    • Paraphrase: All women survivors if all women are fortunate are fortunate.
    • See comment to exercise 12, above.
71. If there are any survivors and only women are survivors, then they are women. Answer
    • ∀X (((s(X)) & ∀Y (s(Y) ⇒ w(Y))) ⇒ w(X))
    • Paraphrase: All survivors if all survivors are women are women.
    • See comment to exercise 12, above.
72. If every position has a future and no employees are lazy, then some employees will be successful. Answer
    • (∀X (p(X) ⇒ f(X)) & ∀Y (e(Y) ⇒ ~l(Y))) ⇒ ∃Z (e(Z) & s(Z))
73. If any employees are lazy, then if some positions have no future, then they will not be successful. Answer
    • ∀X ((e(X) & l(X)) ⇒ (∃Y (p(Y) & ~f(Y)) ⇒ ~s(X)))
    • Paraphrase: All lazy employees if some positions have no future will be unsuccessful.
    • See comment to exercise 12, above.
74. If any employees are lazy and some positions have no future, then some employees will not be successful. Answer
    • (∃X (e(X) & l(X)) & ∃Y (p(Y) & ~f(Y))) ⇒ ∃Z (e(Z) & ~s(Z))
    • Note that this is not an existentially quantified conditional. It is an unquantified conditional. Not even the antecedent is quantified; rather it is an unquantified conjunction whose conjuncts are existentially quantified.
75. If any husband is unsuccessful, then if all wives are ambitious, then some wives will be disappointed. Answer
    • ∃X (h(X) & ~s(X)) ⇒ (∀Y (w(Y) ⇒ a(Y)) ⇒ ∃Z (w(Z) & d(Z)))
    • Note that this is not an existentially quantified conditional; it is an unquantified conditional whose antecedent is existentially quantified.
76. If any husband is unsuccessful, then if some wives are ambitious, he will be unhappy. Answer
    • ∀X ((h(X) & ~s(X)) ⇒ (∃Y (w(Y) & a(Y)) ⇒ u(X)))
    • Paraphrase: All unsuccessful husbands if some wives are ambitious will be unhappy.
    • See comment to exercise 12, above.

    Translations at pp. 128-29

77. Dead men tell no tales. Answer
    • ∀X ((d(X) & m(X)) ⇒ ∀Y (t(Y) ⇒ ~t(X,Y)))
    • This is Copi's answer. It is equivalent to ∀X ∀Y ((m(X) & d(X) & t(Y)) ⇒ ~t(X,Y)), which is in a format I find more natural. (I'll use this format rather than Copi's in the following translations. )
    • Technical aside: the format I prefer is called "prenex normal form". In prenex normal form, all the quantifiers in a formula are stacked at the left end, none is negated, and the scope of each extends to the end of the whole formula.
78. A lawyer who pleads his own case has a fool for a client. Answer
    • ∀X ∀Y ((l(X) & p(X,X)) ⇒ (c(Y,X) & f(Y)))
    • Paraphrase: All lawyers who plead their own cases have fools for clients.
    • Like the English sentence, my translation does not explicitly say that this lawyer is his own client, or that this lawyer is a fool. These propositions follow, of course, but the wit of the original lies in leaving them implicit. At first we think the lawyer and client differ; then we infer that they are the same, and we thereby know who the fool is. We can make these implicit meanings explicit thus: ∀X ((l(X) & p(X,X)) ⇒ (c(X,X) & f(X))).
    • An existential quantifier on 'y' would say: for all lawyers, and at least one client. . . ". It may be true that all lawyers who plead their own cases have at least one foolish client, but this wouldn't guarantee the inference that one of the foolish clients is the lawyer him/herself. This paraphrase of the original shows why the universal quantifier is appropriate: all clients of lawyers who are pleading their own cases are fools.
    • My answer above is not equivalent to ∀X ∀Y ((l(X) & p(X,X) & c(Y,X)) ⇒ f(Y)), but this translation is arguably just as close to the English.
79. A dead lion is more dangerous than a live dog. Answer
    • ∀X ∀Y ((~a(X) & l(X) & a(Y) & d(Y)) ⇒ d(X,Y))
    • Paraphrase: All dead lions are more dangerous than all dead dogs.
80. Uneasy lies the head that wears the crown. Answer
    • ∀X ∃Y ((h(X) & c(Y) & w(X,Y)) ⇒ u(X))
    • Paraphrase: All heads that wear at least one crown lie uneasy.
    • We use an existential quantifier on crowns ('y') because it is not the first quantifier and we mean "some crown or other" rather than "some crown in particular". (See my translation tips on the order of quantifiers, and on "something" for more details. )
81. If a boy tells only lies, none of them will be believed. Answer
    • ∀X ((b(X) & ∀Y (t(X,Y) ⇒ l(Y))) ⇒ ∀Z (t(X,Y) ⇒ ∀U ~b(U,Z)))
    • Paraphrase: For all boys who tell only lies (all boys such that all that they tell are lies), all that they tell is not believed by anybody.
    • I've given Copi's answer above. I prefer the first of the two below, which is equivalent to Copi's. The second is not equivalent to his, but is arguably just as close to the English.
    • ∀X ∀Y ∀Z ∀U ((b(X) & (t(X,Y) ⇒ l(Y))) ⇒ (t(X,Y) ⇒ ~b(U,Z)))
    • ∀X ∀Y ∀Z ∀U (((b(X) & t(X,Y)) ⇒ l(Y)) ⇒ (t(X,Y) ⇒ ~b(U,Z))))
    • Remember that "only A's are B's" is equivalent to "all B's are A's". This is true in polyadic predicate logic, just as in monadic. Massage the English until you can translate it. In this case: ". . . x tells only lies" = "only lies are told by x" = "all that x tells are lies".
82. Anyone who consults a psychiatrist ought to have his head examined. Answer
    • ∀X ∃Y ((p(X) & s(Y) & c(X,Y)) ⇒ o(X))
    • Paraphrase: All persons who consult at least one psychiatrist ought to have their head examined.
    • The quantifier on psychiatrists ('y') is existential because we mean "some psychiatrist or other" rather than "some psychiatrist in particular". See comment to problem 4 above.
    • Copi has made this easy for us by defining o(X) as a jumbo predicate with much internal structure. If we want to make this internal structure explicit, we could define new predicates, h(X), "x is a head", and e(X,Y), "x ought to be examined by y". Replacing o(X) with these finer-grained predicates gives us: ∀X ∀Y ∃Z ((p(X) & s(Y) & c(X,Y) & h(Z)) ⇒ e(Z,Y)).
83. No one ever learns anything unless he teaches it to himself. Answer
    • ∀X ∃Y ((p(X) ⇒ ~l(X,Y)) <~> t(X,Y,X))
    • Copi prefers to translate "unless" as inclusive disjunction: ∀X ∀Y ((p(X) ⇒ ~l(X,Y)) | t(X,Y,X)). I prefer exclusive disjunction (the negation of material equivalence) because we seem to mean that the two propositions cannot both be true. If we learn that the second is true, we will retract our affirmation of the first.
    • The existential quantifier on what is learned ('y') means that we are saying: no one ever learns something or another (at least one thing) unless. . . . If it were universal, we'd be sdaying: no one every learns all things unless. . . .
    • One of the central meanings of this English sentence can be expressed quite simply without the troublesome "unless": ∀X ∃Y ((p(X) & l(X,Y)) ⇒ t(X,Y,X)). This says: All people who know (have learned) at least one thing taught it to themselves.
    • None of these translations captures the "ever" in the English (because Copi didn't give us a predicate to cover it). Let t(X) mean "x is a time". Change l(X,Y) from "x learns y" to l(X,Y,Z), "x learns y at z". Here is how to insert the "ever" into my first translation above: ∀X ∀Y ∀Z (((p(X) & t(Z)) ⇒ ~l(X,Y,Z)) t(X,Y,X))
84. Delilah wore a ring on every finger and had a finger in every pie. Answer
    • ∀X ∃Y ((f(X,d) & r(Y)) ⇒ (o(Y,X)) & ∃X ∀Y ((f(X,d) & p(Y)) ⇒ i(X,Y))
    • Paraphrase: For all of Delilah's fingers, and all pies, there is at least one ring on the fingers and at least one finger in the pies.
    • This is a conjunction of two quantified statements; no quantifier's scope extends over the main conjunction. If we didn't respect this fact, we might try ∀X ∃Y ∀Z ((f(X,d) & r(Y) & p(Z)) ⇒ (o(Y,X) & i(X,Z))). But this says that Delilah had a ring on every finger and every finger (not at least one finger) in every pie. The number of fingers must be requantified from the first clause to the second: in re rings, it is universally quantified, in re pies existentially. We could let the scopes of the ring and pie quantifiers extend over the whole formula provided we requantified on fingers in the middle: ∀X ∃Y ∀Z ((f(X,d) & r(Y) & p(Z)) ⇒ ∃X (o(Y,X) & i(X,Z))).
85. Any man who hates children and dogs cannot be all bad. Answer
    • ∀X ∀Y ∀Z ((m(X) & c(Y) & d(Z) & h(X,Y) & h(X,Z)) ⇒ ~b(X))
    • Paraphrase: No men who hate all children and who hate all dogs are all bad.
    • "All bad" is built into the predicate b(X), so we need not quantify badness.
86. Anyone who accomplishes anything will be envied by everyone. Answer
    • ∀X (p(X) ⇒ (∃Y a(X,Y) ⇒ ∀Z (p(Z) ⇒ e(Z,X))))
    • Paraphrase: All people who accomplish at least one thing will be envied by all people. (Note that it follows that they will envy themselves. )
    • I've given Copi's answer above. Again, I prefer this format: ∀X ∃Y ∀Z ((p(X) & p(Z) & a(X,Y)) ⇒ e(Z,X))).
87. To catch a fish, one must have some bait. Answer
    • ∀X ∃Y ∃Z ((p(X) & f(Y) & b(Z) & c(X,Y)) ⇒ h(X,Z))
    • Paraphrase: All persons who catch at least one fish use at least some bait.
    • Note how the "must" in "must have" is translated by use of the "all. . . do" construction.
88. Every student does some problems, but no student does all of them. Answer
    • ∀X ∃Y ((s(X) & p(Y)) ⇒ d(X,Y)) & ~∃X ∀Y (s(X) & p(Y) & d(X,Y))
    • The second conjunct could also be translated this way: ∀X ∀Y ((s(X) & p(Y)) ⇒ ~d(X,Y))
    • Paraphrase: All students do at least one problem (some problem or other), but it is false that there is a student who does all problems.
89. Any contestant who answers all the questions put to him will win any prize he chooses. Answer
    • ∀X ∀Y ∀Z (((c(X) & q(Y) & p(Z) & p(Y,X)) ⇒ a(X,Y)) ⇒ (c(X,Z) ⇒ w(X,Z)))
    • Paraphrase: For all contestants, if all questions put to them are answered, then if they choose a prize, they win that prize.
    • If we existentially quantified the prize ('z'), we'd be saying that the winner wins some prize or other, at least one prize, not necessarily the one he chooses.
90. Every son has a father, but not every father has a son. Answer
    • ∀X ∃Y ((p(X) & m(X) & p(Y) & m(Y)) ⇒ p(Y,X)) & ∃X ∀Y (p(X) & m(X) & p(Y) & m(Y) & ~p(X,Y))
    • Paraphrase: For all sons, there is a person who is his father, and there is a father who does not have any sons.
    • Copi makes this difficult by giving us an austere dictionary. We have no predicate s(X,Y), "x is the son of y", and no predicate f(X,Y), "x is the father of y". Instead we have predicates for maleness, personhood, and parenthood. We must describe sons as male persons with parents. We must describe fathers as male persons with children.
91. A person is maintaining a nuisance if he has a dog that barks at everyone who visits its owner. Answer
    • ∀X (p(X) ⇒ (∃Y (d(Y) & h(X,Y) & ∀Z ((p(Z) & v(Z,X)) ⇒ b(Y,Z))) ⇒ ∃U (n(U) & m(X,U)))))
    • Paraphrase: All people who have at least one dog that barks at all people who visit its owner maintain a nuisance.
    • I've given Copi's answer above. Again, I prefer this format: ∀X ∃Y ∀Z ∃U ((p(X) & d(Y) & h(X,Y) & p(Z) & v(Z,X) & b(Y,Z)) ⇒ (n(U) & m(X,U)))
92. A doctor has no scruples who treats a patient who has no ailment. Answer
    • ∀X ∃Y ∃Z ∀W ((d(X) & s(Y) & p(Z) & a(W) & t(X,Z) & ~h(Z,W)) ⇒ ~h(X,Y))
    • Paraphrase: No doctors who treat at least one patient who does not have at least one ailment have any scruples.
93. A doctor who treats a person who has every ailment has a job for which no one would envy him. Answer
    • ∀X ∃Y ∀Z ∃W ∀U ((d(X) & p(Y) & t(X,Y) & a(Z) & h(Y,Z) & j(W) & p(U)) ⇒ (h(X,W) & ~e(U,X,W)))
    • Paraphrase: All doctors who treat at least one person who has every ailment have at least one job such that no person envies those doctors that job.
94. If a farmer keeps only hens, none of them will lay eggs that are worth setting. Answer
    • ∀X ∀Y ∀Z ((f(X) & h(Y) & k(X,Y) & e(Z)) ⇒ (l(Y,Z) & ~w(Z)))
    • Paraphrase: All farmers who keep only hens, keep hens that lay eggs that are not worth setting.
    • ". . . x keeps only hens" = "only hens are kept by x" = "all that x keeps are hens".
    • "no y's will lay eggs with property W" = "all y's will lay eggs without property W".

    Translations at p. 129

95. Everyone buys something from some store (or other). Answer
    • ∀X ∃Y ∃Z ((p(X) & s(Y)) ⇒ b(X,Z,Y)).
96. There is a store from which everyone buys something (or other). Answer
    • ∃X ∀Y ∃Z ((s(X) & p(Y)) ⇒ b(Y,Z,X)).
97. Some people make all their purchases from a single store. Answer
    • ∃X ∀Y ∃Z ((p(X) & s(Z)) ⇒ b(X,Y,Z)).
98. No one buys everything it (sic) sells from any store. Answer
    • ∀X ∀Y ∀Z ((p(X) & s(Y)) ⇒ ~b(X,Z,Y)).
    • Paraphrase: No person buys everything whatsoever from any store. This claim is not limited to what a store sells, but refers to everything whatsoever. The English is more particular, but I don't think it can be translated satisfactorily without adding the two-place predicate, s(X,Y), x sells y. With this predicate, I would translate the sentence thus: ∀X ∀Y ∀Z ((p(X) & s(Y) & s(Z,Y)) ⇒ ~b(X,Z,Y)). This translation interprets the English to be saying: No person buys everything a store sells from any store.
99. No one buys things from every store. Answer
    • ∀X ∃Y ∀Z ((p(X) & s(Y)) ⇒ ~b(X,Y,Z)).
100. No store has everyone for a customer. Answer
     • ∀X ∀Y ∃Z ((s(X) & p(Y)) ⇒ ~b(Y,Z,X)).
     • We have no customer predicate, so we describe a customer as a person who buys at least one thing from a store.
101. No store makes all its sales to a single person. Answer
     • ∀X (s(X) ⇒ ∀Y (p(Y) ⇒ ∃Z (∃W b(W,Z,X) & ~b(Y,Z,X))))
     • I've printed Copi's answer above; in prenex normal form it is: ∀X ∀Y ∃Z ∃W ((s(X) & p(Y)) ⇒ (b(W,Z,X) & ~b(Y,Z,X)))).
     • I think Copi is wrong here. His notation asserts: At least one person buys at least one thing from all stores but not everyone buys at least one thing from all stores. Or: All stores have at least one customer but none has all people as customers.
     • My translation: ∀X ∀Y ∃Z ((s(X) & p(Z)) ⇒ ~b(Z,Y,X)).

     Translations at pp. 129-30

102. Nobody donates to every charity. Answer
     • ∀X ∀Y ∃Z ((p(X) & c(Y)) ⇒ ~d(X,Z,Y)).
     • Paraphrase: all people do not donate even one thing to every charity.
103. Nobody donates money to every charity. Answer
     • ∀X ∃Y ∀Z ((p(X) & m(Y) & c(Z)) ⇒ ~d(X,Y,Z)).
     • The quantifier on money ('y') is existential, because we mean: nobody donates some money or other. . . . If it were universal, we'd be saying: noboty donates all money. . .
104. Nobody donates all of his money to charity. Answer
     • ∀X ∀Y ∀Z ((p(X) & m(Y) & b(Y,X) & c(Z)) ⇒ ~d(X,Y,Z)).
105. Nobody donates all of his money to any single charity. Answer
     • ∀X ∀Y ∃Z ((p(X) & m(Y) & b(Y,X) & c(Z)) ⇒ ~d(X,Y,Z)).
106. Nobody donates all of his belongings to any single charity. Answer
     • ∀X ∀Y ∃Z ((p(X) & b(Y,X) & c(Z)) ⇒ ~d(X,Y,Z)).
107. Nobody gives all of his donations to any single charity. Answer
     • ∀X ∀Y ∃Z ((p(X) & c(Z)) ⇒ ~d(X,Y,Z)).
108. No charity receives all of its money from any single person. Answer
     • ∀X ∀Y ∃Z ((c(X) & m(Y) & p(Z) & b(Y,X)) ⇒ ~d(Z,Y,X)).
109. No charity receives all of his (sic) money from any single person. Answer
     • ∀X ∀Y ∃Z ((c(X) & m(Y) & p(Z) & b(Y,Z)) ⇒ ~d(Z,Y,X)).
     • Insofar as this (horrible) sentence is supposed to differ from that in problem 32, I suppose it means this: no charity receives all the money belonging to any single person. That is the sentence I translated above.
110. No charity receives all of his (sic) donations from any single person. Answer
     • ∀X ∀Y ∃Z ((c(X) & p(Z)) ⇒ ~d(Z,Y,X)).
     • Paraphrase: no charity receives all the donations which any single person donates. No charity is the only recipient to which any single person donates.
111. No charity receives all of his (sic) donations from any one donor. Answer
     • ∀X ∀Y ∃Z (c(X) ⇒ ~d(Z,Y,X)).
     • Insofar as this (atrocious) sentence differs from that in problem 34, I suppose it means what 34 means but without the qualification that the donor is a person. No charity receives all the donations which any single donor (who might be a non-person) donates. (Ugh!) That is the sentence I translated above.
112. No charity receives only money as donations. Answer
     • ∃X ∃Y ∀Z (c(Z) & ~m(X) & d(Y,X,Z)).
     • Paraphrase: Not all that is donated to any charity is money. Something which is donated to every charity is not money.
     • Donors are not mentioned in the English, but must be included in the d(X,Y,Z) predicate. The English clearly means that something which is donated by someone or another. . . , rather than something which is donated by everyone. . . . So the donors are existentially quantified.
113. Somebody gives money to charity. Answer
     • ∃X ∃Y ∃Z (p(X) & m(Y) & c(Z) & d(X,Y,Z)).
     • All quantifiers are existential. The first one means "some person in particular". The second and third mean, respectively, "some money or other" and "some charity or other". See my translation tips on quantifier order and on "something".
114. Somebody donates all of his money to charity. Answer
     • ∃X ∀Y ∃Z (p(X) & m(Y) & b(Y,X) & c(Z) & d(X,Y,Z)).
115. At least one person donates all of his belongings to a single charity. Answer
     • ∃X ∃Y ∃Z (p(X) & b(Y,X) & c(Z) & d(X,Y,Z)).
116. At least one person gives all of his donations to a single charity. Answer
     • ∃X ∀Y ∃Z (p(X) & c(Z) & d(X,Y,Z)).
117. Some charities receive donations from everybody. Answer
     • ∃X ∃Y ∀Z (c(X) & p(Z) & d(Z,Y,X)).
     • The English is saying that some charities receive some donation or other from everybody, so the donations ('y') are existentially quantified.
118. Some charities receive donations from every donor. Answer
     • ∃X ∃Y ∀Z (c(X) & d(Z,Y,X)).
     • This is the same as problem 41 except that the donors are not restricted to persons.
119. Some donations are not given to a charity. Answer
     • ∃X ∃Y ∃Z (~c(Z) & d(Y,X,Z)).
     • Paraphrase: There is something which is not a charity which receives some donation or other from some donor or other.
120. Some donors donate to every charity. Answer
     • ∃X ∃Y ∀Z (c(Z) & d(X,Y,Z)).
     • Paraphrase: At least one donor donates something or other to every charity.
121. Every charity receives donations from at least one donor. Answer
     • ∀X ∃Y ∃Z (c(X) ⇒ d(Z,Y,X)).
     • Paraphrase: every charity receives some donation or other from some donor or other.

     Translations at p. 130

122. Whoso sheddeth man's blood, by man shall his blood be shed. Answer
     • ∀X ((p(X) & ∃Y (m(Y) & ∃Z (b(Z) & b(Z,Y) & s(X,Z)))) ⇒ ∃U (m(U) & ∃W (b(W) & b(W,X) & s(U,W))))
     • This is Copi's answer. I prefer this format: ∀X ∃Y ∃Z ∃U ∃W ((p(X) & m(Y) & b(Z) & b(Z,Y) & s(X,Z)) ⇒ (m(U) & b(W) & b(W,X) & s(U,W)))
     • m(X) = x is a man; p(X) = x is a person; b(X) = x is blood; s(X,Y) = x sheds y; b(X,Y) = x belongs to y
     • Paraphrase: For every person who sheds at least some blood belonging to at least one man, there is at least one man who will shed the blood of the (blood-shedding) person. (Technically, the avenging man may be the same as the blood-shedding person, but need not be. )
123. His hand will be against every man, and every man's hand against him. Answer
     • ∃X ∃Y ∀Z ((p(X) & h(Y) & m(Z) & b(Y,X)) ⇒ a(Y,Z)) & ((p(X) & h(Y) & m(Z) & b(Y,Z)) ⇒ a(Y,X))
     • p(X) = x is a person; h(X) = x is a hand; b(X,Y) = x belongs to y; m(X) = x is a man; a(X,Y) = x is against y
     • Paraphrase: There is a person at least one of whose hands is against every man, and at least one hand of every man is against that person.
124. The sun shall not smite thee by day, nor the moon by night. Answer
     • ~s(s,t,d) & ~s(m,t,n)
     • s = the sun; m = the moon; t = thee; d = day; n = night; s(X,Y) = x smites y by z
     • We could avoid 's' as a constant for the sun, and introduce the predicate s(X) = "x is the sun", and so on for m, t, d, and n. We can always replace individuals with suitable predicates because to be a particular individual (e. g. the sun) is equivalent to being an arbitrary individual, x, with the attribute of being that particualr individual (e. g. the attribute of being the sun). But to treat individuals in that way would add complexity to the translation without improving its accuracy.
125. A wise son maketh a glad father. Answer
     • ∀X ∀Y ((m(X) & w(X)) ⇒ (m(X,Y) & f(Y,X) & g(Y)))
     • m(X) = x is male; f(X,Y) = x is y's father; w(X) = x is wise; g(X) = x is glad; m(X,Y) = x maketh y
     • Paraphrase: All wise males make their fathers glad.
126. He that spareth his rod hateth his son. Answer
     • ∀X ((m(X) & ∀Y ((r(Y) & b(Y,X)) ⇒ s(X,Y))) ⇒ ∀Z ((m(Z) & p(X,Z)) ⇒ h(X,Z)))
     • This is Copi's answer. I prefer this format: ∀X ∀Y ∀Z ((m(X) & r(Y) & b(Y,X) & s(X,Y) & m(Z) & p(X,Z)) ⇒ h(X,Z))
     • m(X) = x is male; r(X) = x is a rod; b(X,Y) = x belongs to y; s(X,Y) = x spareth y; h(X,Y) = x hateth y; p(X,Y) = x is the parent of y
     • Paraphrase: All men who spare their rods hate their sons.
127. The borrower is servant to the lender. Answer
     • ∀X ∀Y (b(X,Y) ⇒ s(X,Y))
     • b(X,Y) = x borrows from y; s(X,Y) = x is servant to y
     • Paraphrase: All who borrow are servants to those from whom they borrow. (I tightened up the original, which is vague on the question whether borrowers are servants to their own lenders or to any old lenders. )
128. Whose diggeth a pit shall fall therein: and he that rolleth a stone, it will return upon him. Answer
     • ∀X ∃Y ((p(X) & h(Y) & d(X,Y)) ⇒ f(X,Y)) & (∀X ∃Y (p(X) & s(Y) & r(X,Y)) ⇒ b(Y,X))
     • p(X) = x is a person; h(X) = x is a hole (pit); d(X,Y) = x diggeth y; f(X,Y) = x falls into y; s(X) = x is a stone; r(X,Y) = x rolleth y; b(X,Y) = x returns (back) upon y.
     • Paraphrase: All persons who dig at least one pit will fall into that pit, and all persons who roll at least one stone will have that stone return (back) upon them.
129. The fathers have eaten sour grapes, and the children's teeth are set on edge. Answer
     • ∃X ∃Y ∃Z ∀W ((f(X,Y) & g(Z) & s(Z) & e(X,Z) & t(W)) ⇒ (b(W,Y) & e(W)))
     • f(X,Y) = x is the father of y; g(X) = x is a grape; s(X) = x is sour; e(X,Y) = x eats y; t(X) = x is a tooth; b(X,Y) = x belongs to y; e(X) = x is set on edge
     • Paraphrase: At least some fathers have eaten at least some sour grapes, and all the teeth of the children of those fathers are set on edge.
130. The foxes have holes, and the birds of the air have nests; but the Son of man hath not where to lay his head. Answer
     • ∀X ∃Y ((f(X) & h(Y)) ⇒ h(X,Y)) & ∀X ∃Y ((b(X) & n(Y)) ⇒ h(X,Y)) & ∀X ∃Y ∀Z ((s(X) & c(Y) & b(Y,X)) ⇒ ~p(Z,X,Y))
     • f(X) = x is a fox; h(X) = x is a hole; h(X,Y) = x has y; b(X) = x is a bird of the air; n(X) = x is a nest; s(X) = x is a son of man; c(X) = x is a cranium (head); b(X,Y) = x belongs to y; p(X,Y,Z) = x is a place for y to lay z
     • Paraphrase: All foxes have at least one hole, and all birds of the air have at least one nest, and all the sons of man lack even one place to lay their craniums.
     • We could have packed some of this detail into jumbo predicates, e. g. p(X,Y) = x is a place for y to lay its head.
131. the good that I would, I do not; but the evil which I would not, that I do. Answer
     • ∀X ((g(X) & w(i,X)) ⇒ ~d(i,X)) & ∀Y ((e(Y) & ~w(i,Y)) ⇒ d(i,Y))
     • g(X) = x is a good thing; e(X) = x is an evil thing; i = I; w(X,Y) = x would (wish to) do y; d(X,Y) = x does y
     • Paraphrase: All good things I would wish to do, I do not do, and all evil things I would not wish to do, I do.