Corrections to Logic and Philosophy, 9th Edition, 1st Printing
Paul Tidman/Howard Kahane. ©2003. ISBN: 0534561721.
Page Number
Corrected Text
Inside front cover, left
8. Constructive Dilemma (CD)
pq
pr
qs/rs
Rule UI
1.
(…w…) results from replacing each occurrence of u free in (…u…) with a w that is
free in (…w…) (making no other changes).
Exercise 1-2
In each of numbers 5 and 6 add the word “an” in front of the first occurrence of the word
“argument” so that these items begin as follows: “Is it possible to produce an example of an
argument …?”
Exercise 2-5
9.
Anthony will start at quarterback only if he doesn’t skip practice this afternoon.
(S = “Anthony skips practice this afternoon”)
10.
Anthony is the starting quarterback, unless Chris recovers from his shoulder injury.
(C = “Chris recovers from his shoulder injury”)
7.
(G  ~ O)  (J  L) or G  [~ O  (J  L)]
8.
~ L  [(~ G  ~ J)  O]
9.
(~ L  ~ O)  (G  J)
Exercise 2-7
The answer to #8 is ~M  D not ~M  ~D as given in the answer key.
Examples
1.
~ (A  B)
Exercise 3-2
The answer to #12 is False, not 'indeterminate' as the book says. The antecedent of the
conditional is true because both embedded conditionals have false antecedents and are
therefore true.
Examples (at bottom of page, right column)

B  A
T T T
F T T
T F F
F T F
(at top of page)
…On the other hand, B  A does not logically imply A  B, since on line two B  A is true
and A  B is false.
Inside front cover, right
11
38
42
43/437
46
54/438
60
61
(right column)

B  A
T T T
F T T
T F F
F T F
1/27/04
1
63
69
69/446
73
74
77/447
85
93
In the example, the truth table is incorrect. ~C should be listed as the third premise, and ~A
should be listed as the conclusion. The truth columns are also incorrect, but the second line
of the truth table still constitutes a counterexample to this argument.
(at top of page)
1.
[(A  B)  ~ C]
T T T F F T
Exercise 3-7
The correct answer to #10 is: A,B-True; C,D-False; or D-True; A,B,C-False; or C,D-True;
A,B-False; or B,D-True; A,C-False; or B,C,D-True; A-False.
(last line of page)
~ (p  q)  (r  s) (the “expanded” form).
Examples (#6, Forms column, last two forms)
p  ~ (q  r)
~ p  ~ (q  r)
Exercise 3-12
The truth table given for problem #12 is mistaken on its right hand side—incorrect values are
given for the first "p" following the biconditional, and this percolates up through the rest of
the table. The statement is contingent however.
Exercise 4-2
(8) 3. ~ I / G
The premise on line 5 is missing parentheses and should read ~D  (R  F).
111/452
In the example, line 14 should read L  G, not L  ~G; there's no reason for the negation sign
and it disappears in the next line.
Exercise 4-4
In the answer to problem #8, the "V" in line 2 should be a "U".
Exercise 4-6
(9) 3. Q /  ~ P
The conclusion given in the answer key to problem #2, set 4-2 reads "~A" but should be "A".
117
The second paragraph from the bottom refers to the example on the previous page, p. 116.
119
5.
119
The line numbers in the justification column of the proof are mistaken. For line 6, it should
read "4-5 CP".
The line numbers in the justification column of the proof are mistaken. For line 7 it should
read "3-6 CP"
The line numbers in the justification column of the proof are mistaken. For line 8 it should
read "2-7 CP".
5-1: #8
Left hand parentheses for #8 is missing.
94
95/449
98
119
119
123
124/455
124/455
126
128/457
1/27/04
D
1, 4 MP
The justification for line 11 of exercise #12, set 5-1 reads "4-10 CD" but should read "4-10
CP".
The justification for line 3 of exercise #14, set 5-1 reads "4 Simp" but should read "1, 2 MP".
(top proof)
7.
A
4,6 DS
8.
A~A
1,7 Conj
9.
B
4–8 IP
Exercise 5-4
In problem 4, with IP, the justification for line 11 should be "3, 10 Conj" not "4, 10 Conj".
2
131/459
In the answer to #8 set 5-6, the justifications for lines 12 and 13 should both read “11, Simp”.
134
(In the Examples section, “Syntactic Concepts” and “Semantic Concepts” are reversed.)
136
10. Equivalence Elimination ( E)
pq/pq
pq/qp
(bottom proof)
3.
A
4.
B
5.
B ~ B
6.
~ A
Exercise 5-9
(13) 1. A
138
139
151
159/463-4
168/466
170
173
178/466
178/466
187
189
190-1
190-1
191/469
196
199
1/27/04
AP
1,3  E
2,4  I
3–5 ~ I
/ ~ ~ A
The first sentence of this page says that the negated conditional ~(A  B) is equivalent to the
conjunction ~A  B. This is false; it is equivalent to the conjunction A  ~B.
Exercise 6-3 instructs you to do the trees for exercise 4-11, but should tell you to do the trees
for exercise 4-13, as these are the exercises to which the answers are printed in the back of
the book for 6-3.
Exercise 7-2
The answer for #6 lists F, G, D, K, and H as property constants, but should list only F, D,
and G.
In #9 on the list of examples, the symbolization is given as "(x)Sx  (x)Bx" but should read
"(x)Sx  (x)Bx".
Exercise 7-5
Problem #11 uses the quantifier "most"; but "most" has no symbolization in terms of "(x)" or
"( x)".
Exercise 7-6
In the answer to problem #6, the second "W" should be a "K".
Exercise 7-6
In the answer to problem #10, the symbolized expression is missing a horseshoe and should
read "(x)[(Ax  Rx)  ~Mx]".
(third line from bottom)
Hx = “x is greater than 10”
Exercise 8-2
The conclusion of problem #2, set 8-3, should read "(x)~Bx" instead of "~Bx".
Examples
The truth value assigned to 'Cb' in the premise of the example argument should be "T" not
"F"; as it stands the truth value of '' is computed incorrectly.
Examples
The second conjunct of the conclusion of the example argument is incorrectly described. The
consequent of the conditional should be "Cb" not "Ca".
The answer to problem #2, set 8-3 should contain '' symbols in the first premise and not ''
symbols.
The justification for line 5 of the example should read "3, 4 MP" not "1, 2 MP".
(top line)
Similarly, UG is a valid step (with the exceptions to be noted in the next section)…
3
201
Examples
(2) 12. (x)[Fx  (y)Gy]
3 EG
Exercise 9-1
(1) 3. (x) (Dx Gx) / (x)(~ Dx  Gx)
201/470
201/470
201/470
202/470
205/471
208-9
209
210
212/471
213-4
217
217
221
225/473
226/473
226/473
226/473
226/473
227/473
228
1/27/04
In the solution to #2, problem set 9-1, there should be no line number next to "AP" in the
justification for line 5.
In the solution to #2, problem set 9-1, there is no CP box, but there should be (extending to
the top of line 5)
In the solution to #2, problem set 9-1, the last step of the proof is not given; the line "12. (
x)(Dx  Ax) 11, EG" should be added.
The last line of the answer to problem #6 should read "9. (y)By 8, UI" not "9. ( y)By 8,
EG".
Exercise 9-2
(2) 4. (x)(~ Gx  ~ Mx)  Fa
1 Contra
On pg. 209, line 3 the book says "…in line 4 [the x variables] are changed into a free x and a
free y", but should say "…in line 4 they are changed into a free x and a free z".
The line numbers used in the justifications of lines 3 and 4 of the example are reversed; the
justifications should read "1 EI" and "2 UI", respectively.
In the example that begins at the bottom of page 209, "2 UI" is invoked in both the 'valid' and
'invalid' derivations, when "1, UI" is the proper rule to use in both cases.
Exercise 9-3
In #6, Premise 3 should be “(z)(~Rz  Pz)”
In line 5 of the example in the main text (unshaded) the justification should read "2, 4 MP"
not "2, 3 MP".
The conclusion of problem #12 should read “(x)(Ix  ~Fx)” not “(x)(x  ~Fx)”.
Exercise 9-5
In #16, Premise 2 should be “(x)~(Px  ~Qx)”
In #20, set 10-1, 'sexually' should be omitted, since this adverb was not used in previous
sentences that receive the same dictionary.
Exercise 10-3
The answer to problem #6 is missing a parenthesis and should read
“(x){(Gx  Sx)  (y)[(Dy  Cy)  Txy]}”.
Exercise 10-3
The answer to problem #10 has mismatched parentheses and should read
“(x){Sx  (y)[(Cy  Tjy)  Txy]}”.
Exercise 10-3
The answer to problem #16 contains multiple errors regarding its parentheses and should read
“(x)[(Gx  Sx)  (y)[(By  Cy)  ~Txy]]”.
Exercise 10-3
The answer to problem #18 has mismatched parentheses and should read
“(x){(Ex  ~Pmx)  (y)[(Ey  Tjy)  Dxy]}”.
Exercise 10-3
The answer to problem #30 has a misplaced parentheses and should read:
“(x){[Cx  ~(y)(Cy  Pjy)  (Dxy  Ixy)]  ~Tmx}”
Exercise 10-4
The answer to #6 uses Txy in its symbolization but should use Axy.
The example in the middle of the page should read "(x)[Nx  (y)(Ry  Ayx)]".
4
228
229
229
232
233
234
234
237
238
240
242
244
(second to last paragraph, second to last sentence)
But perhaps this strangeness can be dispelled to some extent by pointing out that
(x)(Nx)  (y)(Ry  Ay) is equivalent to (x)[Nx  (y)(Ry  Ay)].
In the example, seven lines down from the top of the page, the antecedent of the conditional
should read "(x)(Nx)".
In the example, seven lines down from the top of the page, there should be no '(z)' quantifier;
that should simply be omitted.
3.
(x)(Px  ~ Bxg)
Problem #1, set 10-8 presents a valid argument, not an invalid one as the exercise
presupposes.
(top proof)
6.
Lxy
4 UI
Examples (top proof)
2.
(y)(Fx  Fxy)
1 EI
Examples (bottom proof)
In the second example of valid UI/EI applications in the shaded box, the line numbers in the
justifications are off by one; they should read "1, UI" and "2, EI".
In the example in the center of the page, the justification for lines 2-6 is given as "2, UG". It
should in every case be "1, UG".
The justification for line 3 of example #2 in the shaded box is entirely mistaken. Contrary to
what it says, every x is replaced by a z; the use of EI is invalid because z is a bound variable
in the original statement within the scope of ( x).
(third proof)
7.
Gzw
4,6 MP
Exercise 10-11
(1) 2. (x)[Ax  (y)(y  ~ Lxy)]
/ (x)(Qx  ~ x)
(bottom proof)
7.
[(x)Fx  ~ (x) ~ Fx] ~ (x) ~ Fx  (x)Fx]
3,6 Conj
259
In case (2), the line 7, which the justification for line 9 apparently refers to, is missing.
259
In case (4), the line 1, which the justification for line 1 apparently refers to, is missing.
281
In example 3, there should be a dictionary entry “Ix = ‘x is intelligent’”.
281
In example 5, there should be a dictionary entry “Fxy = ‘x is the father of y’”.
437
Exercise 2-7
16.
(~ M S) ~ B
18.
(~ M ~ B)  (~ S  F)
Exercise 3-2
12.
False. The antecedent is true, because it is a conjunction of two true conditionals.
These two conditionals are true because their antecedents are each false.
Exercise 3-10
14.
a, b, i, j
Exercise 3-11
6.
p, p  q, p  ~ q, (p q)  r, (p q)  ~ r, (p  q)  ~ q
12.
p, ~ p, ~ (p  q), ~ [(p q)  r], ~ [p  (q  r)], ~ [(p q)  (r  q)],
~ [(p  q)  (r  s)]
438
446
447
1/27/04
5
452
456
460–1
1/27/04
Exercise 4-11
(8) 3. C  (A  B)
Exercise 5-3
(10) …
3.
~ (C  D)
4.
~ (~ C  D)
5.
~~C~D
6.
~~C
7.
C
8.
D  ~ (A  B)
9.
~D
10.
~ (A  B)
11.
~A~B
12.
~A
13.
~B
14.
B
15.
B~B
16.
CD
17.
~D~C
1 Comm
AP
3 Imp
4 Dem
5 Simp
6 DN
1,7 MP
5 Simp
8,9 DS
10 Dem
11 Simp
11 Simp
2,12 MP
13, 14 Conj
3-15 IP
16 Contra
Exercise 5-9
(Replace all instances of “int” with “I”, and all instances of “elim” with “E”.)
6