Problem 3-2
Part I:
S(1): If P is TRUE and Q = ~~P, Q is also TRUE by the Law of Double Negation.
In general, let us assume that for an arbitrary number k, if Q = P preceded by 2k ~'s, Q will also be TRUE.
Now all we have to do is to prove that for the next case (k+1), Q will be FALSE and that for (k+2), Q will
be true.
Well, if a P preceded by k ~'s is TRUE, P preceded by k+1 ~'s will be FALSE, by the definition of
negation. By the same rule, adding another ~ will again flip the truth value, thus making k+2 TRUE.
So, for all cases where Q = P preceded by an even number of ~'s (k, k+2, K+4, …), if P is true, Q will be
true as well. And, for all cases where Q = P preceded by an odd number of ~'s, if P is TRUE, Q will be
false.
Part II:
Since all whole numbers are even or odd and since all literals (P or ~P, if P is atomic) must be either true or
false, the above example exhausts all cases. Thus, no matter how many negation symbols Q has, it will
have the same truth value as a literal, namely either the literal P or the literal ~P.
Problem 3-9
1. (A  B)  A
AB
2.
3.
(B  (A  B  C))
ABC
Idempotence of 
Idempotence of 
(A  B)  (C  D)  A
(A  B)  A  (C  D)
(A  B  A)  (C  D)
(A  B)  (C  D)
Commutativity of 
Associativity of  (Unnecessary)
4.
(A  B)  (B  C)
(A  B  B)  C
Associativity of  (Unnecessary)
A  B  C
Idempotence of 
5.
(A  B)  C  (B  A)  A
(A  B)  C  (A  B)  A
(A  B)  C  A
Commutativity of 
Idempotence of 
Problem 3-10
1. (Home(Carl)  Home(Claire))
Home(Carl)  Home(Claire)
2.
3.
DeM
[Happy(Max)  (Likes(Carl,Claire)  Likes(Claire,Carl))]
Happy(Max)  (Likes(Carl,Claire)  Likes(Claire,Carl))
Happy(Max)  (Likes(Carl,Claire  Likes(Claire,Carl))
DeM
DeM
[(Home(Max)  Home(Carl))  (Happy(Max)  Happy(Carl))]
[(Home(Max)  Home(Carl))  (Happy(Max)  Happy(Carl))] DN
(Home(Max)  Home(Carl))  (Happy(Max)  Happy(Carl)) DeM
Home(Max)  Home(Carl)  Happy(Max)  Happy(Carl) DeM
Problem 3-15
1. Max is a student, not a disk.
Student(Max)  Disk(Max)
2.
Claire erased Folly at 2pm and then gave it to Max.
Erased(Claire, Folly, 2:00)  Gave(Claire, Folly, Max, 2:00 < t)
Note that this problem is rather bogus (as is the solution) since we don't have any way of specifying an
unspecified time that is later than another time. A function symbol would do the trick, but we don't
have one. It would be nice if they had provided Later(t), for "a time later than t".
Given what we have to work with, this might be a good translation, despite the fact that it has more
precision than the sentence we're given:
Erased(Claire, Folly, 2:00)  Gave(Claire, Folly, Max, 2:01)
3.
Folly belonged to either Max or Claire at 2:05 pm.
Owned(Max, Folly, 2:05)  Owned(Claire, Folly, 2:05)
4.
Neither Max nor Claire erased Folly at 2 pm or at 2:05 pm.
Erased(Max, Folly, 2:00)  Erased(Max, Folly, 2:05)  Erased(Claire, Folly, 2:00) 
Erased(Claire, Folly, 2:05)
5.
2:00 pm is between 1:55 pm and 2:05 pm.
(1:55 < 2:00)  (2:00 < 2:05)
6.
When Max gave Folly to Claire at 2pm , it wasn't blank, but it was five minutes later.
Gave(Max, Folly, Claire, 2:00)  Blank(Folly, 2:00)  Blank(Folly, 2:05)
Problem 3-16
1. Student(Claire)  Student(Max)
Claire is a student, but Max is not.
2.
Disk(Silly)  Owned(Max, Silly, 2:00)
Silly is a disk but Max didn't own it at 2:00 pm.
3.
Owned(Claire, Silly, 2:00)  Owned(Claire, Folly, 2:00)
Claire owned either Silly or Folly at 2:00 pm.
4.
(Erased(Max, Silly, 2:00)  Erased(Max, Folly, 2:00)
Max erased neither Silly nor Folly at 2:00pm.
5.
((Gave(Max, Silly, Claire, 2:00)  Blank(Silly, 2:00)) 
(Gave(Max, Folly, Claire, 2:00)  Blank(Folly, 2:00))) 
Angry(Claire, 2:05)
Either Max gave Silly to Claire at 2:00pm and it was blank at 2:00pm or Max gave Folly to Claire at
2:00pm and it was blank at 2:00pm, and Claire was Angry at 2:05pm.
Problem 3-17
1.
English
The disease known as AIDS
FOL
AIDS (Name)
The disease known as influenza
X is less contagious than y
X is more deadly than y
Flu (Name)
LessContagious(x, y) (Predicate)
MoreDeadly(x, y) (Predicate)
LessContagious(AIDS, Flu)  MoreDeadly(AIDS, Flu)
2.
English
Abe
Stephen
Monday
Sunday
X Fooled y on d (a day)
FOL
Abe (Name)
Stephen (Name)
Monday (Name)
Sunday (Name)
Fooled(x, y, d)
Fooled(Abe, Stephen, Sunday)  Fooled(Abe, Stephen, Monday)
3.
English
Dan
George
Al
Bill
X Admires y and z
FOL
Dan (Name)
George (Name)
Al (Name)
Bill (Name)
Admires(x, y, z) (Predicate)
Admires(Dan, Al, Bill)  Admires(George, Al, Bill)
4.
English
Daisy
Dee
X is a river
X is a jolly miller
X lives on y (a place)
FOL
Daisy (Name)
Dee (Name)
River(x) (Predicate)
JollyMiller(x) (Predicate)
Lives(x, y) (Predicate)
JollyMiller(Daisy)  Lives(Daisy, River(Dee))
3-18
1. (A  B)  (A  B)
Since, by DeMorgan's theorem, (A  B) is (A  B), we have
(A  B)  (A  B)
This is a tautology since it is of the form P  P, and whenever P (A  B) is true, P ((A  B)) must be
false. Because a disjunction is true whenever one side is true and because one side is always true, this
statement is always true. Therefore it is a tautology.
2.
(A  B)  (A  B)
This is not a tautology because of lines 3 & 4.
(
A
T
T

T
F
B )
T
F

T
T
( A
T
T

F
T
B )
F
T
F
F
3.
F
F
T
F
F
F
F
F
F
F
F
T
(A  B)  C
This is not a tautology because of line 2.

F
F
T
T
T
T
T
T
4.

T
T
F
F
F
F
F
F
( A
T
T
T
T
F
F
F
F

T
F
T
T
T
T
T
T
B )
T
T
F
F
T
T
F
F
C
T
F
T
F
T
F
T
F
(A  B)  (A  (B  C))
This is not a tautology because of lines 3-6.
( A
T
T
T
T
F
F
F
F

T
T
F
F
F
F
F
F
B )
T
T
F
F
T
T
F
F

T
T
F
F
F
T
T
T

F
F
F
F
F
T
T
T
(

T
T
T
T
T
F
F
F
A
T
T
T
T
F
F
F
F

T
F
F
F
T
F
F
F
( B
T
T
F
F
T
T
F
F
C )
T
F
T
F
T
F
T
F
3-19
We do this problem by deciding whether any of the lines in the truth table are spurious. What this means is
that if a given statement is necessarily true, assigning the value to false to it would be bogus. So, what we
do is blank out the rows that contain spurious values and then assess the remaining rows. (I'll shade the
spurious rows)
1-1
(A  B)  (A  B)
( A

T
T
T
F
F
F
F
F
B )
T
F
T
F

T
T
T
T
( A
F
F
T
T

F
T
T
T
B )
F
T
F
T
Because this was a tautology anyway, eliminating the spurious rows doesn't make any difference. It's still a
tautology, which means that is logically true
1-2
( A
T
T
F
F

T
F
F
F
B )
T
F
T
F

T
T
F
F
( A
T
T
F
F

F
T
F
F
B )
F
T
F
T
Because only T's are left under the major connective, this turns out to be logically true.
1-3

F
F
T
T
T
T
T
T

T
T
F
F
F
F
F
F
( A
T
T
T
T
F
F
F
F

T
F
T
T
T
T
T
T
B )
T
T
F
F
T
T
F
F
C
T
F
T
F
T
F
T
F
This turns out to be satisfiable (not logically true) because of the F in row 2.

T
T
F
F
F
F
F
F
( A
T
T
T
T
F
F
F
F
B )
T
T
F
F
T
T
F
F

T
T
F
F
F
T
T
T

F
F
F
F
F
T
T
T
(

T
T
T
T
T
F
F
F
A
T
T
T
T
F
F
F
F

T
F
F
F
T
F
F
F
( B
T
T
F
F
T
T
F
F
C )
T
F
T
F
T
F
T
F
This turns out to be satisfiable (not logically true) because of the F's in rows 3 & 4.
2-1
( A
T
T
F
F

T
F
F
F
( A
F
F
T
T

T
T
T
T
B )
T
F
T
F

F
T
T
T
B )
F
T
F
T

F
T
F
F
B )
F
T
F
T
This is still logically true because it was a tautology.
2-2
( A
T
T
F
F

T
F
F
F

T
T
F
F
B )
T
F
T
F
( A
T
T
F
F
This turns out to be satisfiable if a is necessarily false.
2-3

F
F
T
T
T
T
( A
T
T
T
T
F
F

T
T
F
F
F
F
B )
T
T
F
F
T
T

T
F
T
T
T
T
C
T
F
T
F
T
F
T
T
F
F
F
F
F
F
T
F
T
T
This turns out to be logically true because after eliminating the first 4 rows because they are spurious if A is
necessarily false, all we have left are T's.
2-4
( A
T
T
T
T
F
F
F
F

T
T
F
F
F
F
F
F
B )
T
T
F
F
T
T
F
F

T
T
F
F
F
T
T
T

F
F
F
F
F
T
T
T
(
A
T
T
T
T
F
F
F
F

T
T
T
T
T
F
F
F
( B
T
T
F
F
T
T
F
F

T
F
F
F
T
F
F
F
C )
T
F
T
F
T
F
T
F
This turns out to be satisfiable.
3-1
Since there is no C, the condition does not apply.
3-2
Since there is no C, the condition does not apply.
3-3

F
F
T
T
T
T
T
T

T
T
F
F
F
F
F
F
( A
T
T
T
T
F
F
F
F

T
F
T
T
T
T
T
T
B )
T
T
F
F
T
T
F
F
C
T
F
T
F
T
F
T
F
This turns out to be logically true, because when we eliminate row 2 (when A and B are true, C must be
true), we are left with all T's.
3-4
( A
T
T
T
T
F
F
F
F

T
T
F
F
F
F
F
F
B )
T
T
F
F
T
T
F
F

T
T
F
F
F
T
T
T

F
F
F
F
F
T
T
T
(
A
T
T
T
T
F
F
F
F
This continues to be satisfiable, even when A and B imply C.
Problem 20

T
T
T
T
T
F
F
F
( B
T
T
F
F
T
T
F
F

T
F
F
F
T
F
F
F
C )
T
F
T
F
T
F
T
F
1.
2.
Every satisfiable sentence is tt-satisfiable because in order to be satisfiable, at least one row must be
true, which is the definition of tt-satisfiable.
Cube(a)  Tet(a)
This sentence is not satisfiable because a can't be a cube and a tetrahedron at the same time. It is ttsatisfiable because it there is at least one row in its truth table (though spurious), which has the truth
value of T.
3.
Every tt-satisfiable sentence is satisfiable if the atomic sentences of the langauge are logically
independent because logical independence is a condition of all logically true statements being
taugologies. That is, if all statements of the language are logically independent, the only truth tables
that will be logically true are tautologies.
4.
S is tt-satisfiable IFF S is not a tautology because if S were a tautology, S would be unsatisfiable.
In order to be tt-satisfiable, a sentence must be satisfiable.
Problem 21
(The spurious rows will be grayed out.)
1.
Ab=b
( A
T
T
T
T
F
F
F
F
2.

T
T
F
F
F
F
F
F
B )
T
T
F
F
T
T
F
F

T
T
F
T
F
T
F
T
C
F
T
F
T
F
T
F
T

T
T
F
F
F
F
F
F
B )
T
T
F
F
T
T
F
F

T
T
F
T
F
T
F
T
C
F
T
F
T
F
T
F
T
A  Tet(b)
( A
T
T
T
T
F
F
F
F
This is because A and B can't be true at the same time if A  Tet(b) and B  Cube(b), though they can be
false at the same time.
3.
A  Cube(a), B  Tet(a), C  Dodec(a)
( A
T
T
T
T
F

T
T
F
F
F
B )
T
T
F
F
T

T
T
F
T
F
C
F
T
F
T
F
F
F
F
F
F
F
T
F
F
T
F
T
T
F
T
The gray rows are spurious because no two of the statements can be true at the same time.
4.
Cbb
( A
T
T
T
T
F
F
F
F

T
T
F
F
F
F
F
F
C
F
T
F
T
F
T
F
T

T
T
F
T
F
T
F
T
B )
T
T
F
F
T
T
F
F
Any row in which C is true will be spurious, therefore any row in which C is false is spurious.
5.
A  Larger(a, b), B  Larger(b, c), C  Larger(c, a)

T
T
F
F
F
F
F
F
( A
T
T
T
T
F
F
F
F
C
F
T
F
T
F
T
F
T

T
T
F
T
F
T
F
T
B )
T
T
F
F
T
T
F
F
In any row where A and B are true, C must be false. Likewise, if A & C are true, B must be false. And, if
B and C are true, A must be false.
Problem 22
1.

T
T
T
T

F
F
F
F
( B
T
T
F
F

F
F
F
T

T
T
T
F
( C
T
F
T
F
B )
T
T
F
F
This sentence is logically true (because it is a tautology).
2.
A
T
T
T
T
F
F
F
F

T
T
T
T
F
F
F
F

F
F
T
F
F
F
F
F
( B
T
T
F
F
T
T
F
F

T
T
F
T
T
T
T
T

F
T
F
T
T
T
T
T
( C
T
F
T
F
T
F
T
F

T
F
T
F
F
F
F
F
A) )
T
T
T
T
F
F
F
F
This sentence is satisfiable.
3.

F
F
F
F
F
F
F
F
[( 
F
F
F
F
T
T
T
T

F
T
T
T
T
T
T
T
A
T
T
T
T
F
F
F
F

F
T
T
T
F
T
T
T
(B
T
T
F
F
T
T
F
F

T
F
F
F
T
F
F
F

T
T
T
T
T
T
T
T
C))
T
F
T
F
T
F
T
F

T
T
F
F
F
F
F
F
(A
T
T
T
T
F
F
F
F
B )]
T
T
F
F
T
T
F
F
This sentence is unsatisfiable.
4.

T
F
F
F
T
T
T
T
T
F
F
F
T
F
F
F
[(
F
F
F
F
F
F
F
F
T
T
T
T
T
T
T
T
A
T
T
T
T
T
T
T
T
F
F
F
F
F
F
F
F

T
T
T
T
F
F
F
F
T
T
T
T
T
T
T
T
B)
T
T
T
T
F
F
F
F
T
T
T
T
F
F
F
F

F
T
T
T
F
F
F
F
F
T
T
T
F
T
T
T

F
T
T
T
F
T
T
T
F
T
T
T
F
T
T
T
(C
T
T
F
F
T
T
F
F
T
T
F
F
T
T
F
F

T
F
F
F
T
F
F
F
T
F
F
F
T
F
F
F
D)]
T
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
This sentence is satisfiable.
Problem 23
1.
Happy(Carl)
T
T
T
T
F
F
F
F

F
F
F
T
F
F
F
F

F
F
F
T
F
F
F
T
(Home(Carl)
T
T
F
F
T
T
F
F

F
(Carl=Carl
T

T
T
T
F
T
T
T
F
Home(Claire))
T
F
T
F
T
F
T
F
2.
(The shaded rows are crossed out.)
Happy(Carl)
T

T

T
Claire=Claire)
T
T
T
T
F
T
T
T
F
F
T
T
T
F
F
T
F
F
T
T
F
F
T
T
F
T
F
T
F
F
T
F
T
F
F
T
This sentence is unsatisfiable because after eliminating all spurious rows all rows are F.
3.
[Happy(Carl)
T
T
F
F

T
T
T
T

F
F
T
T

T
T
T
T
Happy(Carl)]
T
T
F
F
F
T
F
T
F
T
F
Home(Max)
T
F
T
F
This sentence is logically true because it is a tautology.
4.
[Happy(Carl)
T
T
T
T
F
F
F
F

T
T
T
T
T
T
F
F

T
T
T
T
T
T
T
F
Unhappy(Carl)]
T
T
F
F
T
T
F
F
Home(Max)
T
F
T
F
T
F
T
F
Carl can't be happy and unhappy at the same time, though he can be neither (if he is asleep). This sentence
is satisfiable.
Problem 24
Sentence
1
2
3
4
5
6
7
8
9
10
Tarski's World
Yes
NO
NO
NO
YES
NO
NO
NO
YES
NO
Truth assignment
T T T F, NOT SPURIOUS
T T T, NOT SPURIOUS
F F T T, NOT SPURIOUS
Problem 25
1.
( Cube(a)
T
T
T
T
F

T
T
T
T
T
Cube(b) )
T
T
F
F
T

T
F
T
F
T
Tet(a)
T
F
T
F
T
Cube(b)
T
T
F
F
T
F
F
F
T
F
F
T
F
F
F
T
F
F
F
F
T
F
F
Cube(b) is a tautological consequence of the expression.
2.
Cube(b) is also a logical consequence of the expression.
3.
( Cube(a)
T
T
F
F

T
T
T
F
Cube(b))
T
F
T
F

F
F
T
F

F
F
T
T
Cube(a)
T
T
F
F
Cube(b)
T
F
T
F
Cube(b) is a tautological consequence of the expression.
4. Cube(b) is also a logical consequence of the expression.
5.
Dodec(b)
T
T
T
T
F
F
F
F

T
T
F
F
F
F
F
F
c=b
T
T
F
F
T
T
F
F
Dodec(c)
T
F
T
F
T
F
T
F
Dodec(c) is not a tautological consequence of the expression.
6.
Dodec(c) is a logical consequence of the expression.
7.
Dodec(b)
T
T
T
T
F
F
F
F

T
T
F
F
F
F
F
F
e=c
T
T
F
F
T
T
F
F
Dodec(c)
T
F
T
F
T
F
T
F
Dodec(c) is not a logical consequence of the expression.
Problem 26
If a complex sentence composed of atomic sentences A, B, and C is a tautology, then it will remain a
tautology even if A is replaced with a complex sentence P because in order for the initial sentence to be a
tautology, it has to consider all possible combinations of truth value. Even if we replace A with P, we still
consider all possible combinations, so the situation with respect to tautology remains unchanged.
However, substitution will not necessarily preserve logical truth because B or C could be contained in or
otherwise dependent upon sentences that make up the complex sentence P. If this is the case, one or more
rows may turn out to be spurious.