Translation Tips—Chapter 5 Copi
I. You will have translations like set II pp. 128-130 on the final exam. Your general approach to
translation should be:
 Step 1: Read the dictionary give FIRST. To make it easier on the final, I will have JUST
ONE DICTIONARY FOR ALL THE PROBLEMS.
 Step 2: then read the statement to be symbolized carefully and closely.
 Step 3: Focus on the subject, and write pseudo-code accordingly. Watch out for the
scope of the main quantifier.
 Step 4: Now turn to the components, and expand on them, focusing on only one
quantifier at a time.
 Repeat step 4 until all the structure is in pseudo-code
 Then transcribe into PL notation.
II. Remember your logical equivalences:
 You can use QN now. So if you semantically interpret “no S are P” as “there are no
things that are both S and P”, you can translate it as (x)(Sx  Px); but if it is clearer as
“All S things are non-P things”, then translate it as (x)(Sx  Px).
 Review the other logical equivalences pp. 110-113
 Remember the variables are all now usable: u, v, w, x, y, z. You can select them as you
wish. I personally like to choose different variables for clarity, but as long as you pay
attention to scope, you can often use the same variable. For example:
1.
(x)[Ax v (y)By]
2.
(x)[Ax v (x)Bx]
Both are equivalent, but I find #1 clearer.
 BUT REMEMBER: YOU CAN ONLY USE VARIBLES AS PARTS OF
QUANTIFIERS. E.g., (b) and (b) are ungrammatical.
 You can use {,}, <, >, [, ], (, ) in various orders as you please.
 If you have two dots (conjunctions) or two wedges (two disjunctions), you need not use
brackets internally. “Tolerable ambiguity”…
Examples: (I pick some of the more interesting problems)
#16:
Step 3: For all x, if x is a doctor and x treats a patient with no ailments, then x has no scruples.
Step 4: For all x, if x is a doctor and there exists a y such that y is a patient and x treats y, but y
has no ailments, then x has no scruples.
Step5: For all x, if x is a doctor and there exists a y such that y is a patient and x treats y but it is
not the case that there exists a w such that w is an ailment and y has w, then x has no scruples.
Step 6: For all x, if x is a doctor and there exists a y such that y is a patient and x treats y but it is
not the case has there exists a w such that w is an ailment and y has w, then it is not the case that
there exists a z such that z is a scruple and x has z.
Transcription:
(x){ <Dx  (y)[Py Txy  (w)(Aw  Hyw)] >  (z)(Sz  Hxz)}
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#18:
Step 3: For all u, if u is a farmer and u keeps only hens, then none of those hens will lay eggs that
are worth setting.
Step 4: For all u, if u is a farmer and for any w, u keeps w only if w is a hen, then none of those
hens will lay eggs worth setting.
Step 5: For all u, if u is a farmer and for any w, u keeps w only if w is a hen, then if x is a hen
and u keeps x, then x will not lay eggs worth setting.
Step 6: For all u, if is u is a farmer and for any w, u keeps w only if w is a hen, then for any x, if
x is a hen and u keeps x, then it is not the case that there exists a z such that z is an egg and x
lays z, and z is worth setting.
Transcription:
(u){Fu  < (w)(Kuw  Hw)  [ (x)[(Hx  Kux)  (z)[Ez  Lxz  Wz]] > }
#22:
Step 3: For all x, if x is a person, then x does not buy everything that any given store sells.
Step 4: For all x, if x is a person, then for all y, if y is a store, then x doesn’t buy all of what y
sells.
Step 5: For all x, if x is a person, then for all y, if y is a store, then there exists a z such that x
doesn’t buy z from y.
Transcription:
(x){Px  (y)[Sy  (z)Bxzy]}
OR: (x)(y)[ (Px  Sy)  (z)Bxzy]
OR: (x)(y)[ (Px  Sy)  (z)Bxzy]
#28:
Step 3: For all x, if x is a person, x does not give all his money to charity.
Step 4: For all x, if x is a person, it is not the case that for all y, if y is money and y belongs to x,
then x donates y to charity.
Step 5: For all x, if x is a person, it is not the case that for all y, if y is money and y belongs to x,
then there exists some z, such that z is a charity and x donates y to z.
Transcription:
(x){Px  (y)[(My  Byx)  (z)(Cz  Dxyz)]}
OR: (x){Px  (y)[(My  Byx)  (z)(Cz  Dxyz)]}
OR: (x){Px  (y)[(My  Byx)  (z)(Cz  Dxyz)]}
#44:
Step 3: There exists a v such that v is a donor and v donates to every charity.
Step 4: There exists a v such that v is a person and there exists an x and there exists a y such that
x is a charity and v donates y to x, and v donates to every charity.
Step 5: There exists a v such that v is a person and there exists an x and there exists a y such that
x is a charity and v donates y to x, and for all w, if w is a charity, then there exists a u such that v
donates u to w.
Transcribe:
(v){Pv  (x)(y)(Cx  Dvyx) (w)[Cw  (u)Dvuw}
I would allow you to drop the second interior conjunct—it seems to me to be redundant….
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