Symbolic Logic Problems

Transcription

Symbolic Logic Study Guide: Homework Solutions
67
SECTION 2: HOMEWORK SOLUTIONS
This section includes solutions to the homework problems in the course. Note that due to the nature of symbolic logic,
there are many problems that can have multiple solutions, especially some of the world building problems in Tarski’s
World and several of the translation problems later in the course. Thus, these solutions are possible solutions to the
problems, but there may be others. For some problems, such as proofs, there are multiple correct solutions presented
here.
There are also several Tarski’s World problems that require you to evaluate the truth value of a set of sentences; since
these problems are designed to increase your understanding of the material and you can find the truth values in Tarski’s
World, they are not presented here. They are labeled as “Tarski’s World Drills.”
Problems are numbered c-p, where c is the chapter number and p is the problem number.
For Tarski’s World problems where discerning the size of objects is important, the blocks are labeled as S (small), M
(medium), or L (large).
2.1. Chapter 1 Solutions
There are no homework problems in this chapter. Instead, read the chapter to get an introduction to what the course will
entail as why logic is useful.
Problem 2-1: Tarski’s World Drill
Problem 2-2: Solution looks identical to what is presented in the text.
Problem 2-3: 2-3.wld:
S
Problem 2-4: 2-4.sen:
M
L
68
Problem 2-4: continued, 2-4.wld:
L
M
S
L
S
Problem 2-5: 2-5.wld
M
L
M
S
S
S
Problem 2-9:
Given the functional language and the relational language as follows:
the relational language
the functional language
Names
Claire, Melanie, Jon
Claire, Melanie, Jon
Predicates
TallerThan(x, y)
FatherOf (x, y): x is the father of y.
x=y
TallerThan(x, y)
x=y
Functions
Translations
1. FatherOf(Jon, Claire)
2. FatherOf(Jon, Melanie)
3. TallerThan(Melanie, Claire)
father (x): the father of x
Jon = father(Claire)
Jon = father(Melanie)
TallerThan(Melanie, Claire)
Which can be translated into atomic sentences of relational language?
1. YES – FatherOf(Jon, Melanie)
2. NO – There are no function symbols, connectives, or quantifiers in the relational language. Unless an object is
named, it is not possible to refer to it. To express this sentence, it is necessary to refer to father(Melanie) and
father(Claire), which is not possible.
3. NO – There are no function symbols, connectives, or quantifiers in the relational language. Unless an object is
named, it is not possible to refer to it. To express this sentence, it is necessary to refer to father(Claire) and
father(Jon), which is not possible.
69
Problem 2-10:
One could use the following translation manual:
English
FOL
Names
Carl, Sam, Mary
the same
Predicates
x is the same as y
x is greater than y
x=y
x>y
Functions
the height of x
height (x)
Some sample translations are:
• Carl is taller than Sam.
• Sam and Mary are the same height.
• Mary is shorter than Carl.
height(Carl) > height(Sam)
height(Sam) = height(Mary)
height(Carl) > height(Mary)
Problem 2-14:
Given the following two first order languages:
Language 1
English
Max
Max
Claire
Claire
Names
Predicates
x gave Scruffy to y
x gave Carl to y
Language 2
FOL
GaveScruffy(x, y)
GaveCarl(x, y)
English
Max
Claire
Scruffy
Carl
x gave y to z
FOL
Max
Claire
Scruffy
Carl
Gave(x, y, z)
Functions
1. In the first language:
GaveScruffy(Max, Max)
GaveScruffy(Max, Claire)
GaveScruffy(Claire, Max)
GaveScruffy(Claire, Claire)
2. In the second language
3.
GaveCarl(Max, Max)
GaveCarl(Max, Claire)
GaveCarl(Claire, Max)
GaveCarl(Claire, Claire)
Gave(Max, Max, Max)
Gave(Max, Claire, Max)
Gave(Max, Scruffy, Max)
Gave(Max, Carl, Max)
Gave(Max, Max, Claire)
Gave(Max, Claire, Claire)
Gave(Max, Scruffy, Claire)
Gave(Max, Carl, Claire)
Gave(Max, Max, Scruffy)
Gave(Max, Claire, Scruffy)
Gave(Max, Scruffy, Scruffy)
Gave(Max, Carl, Scruffy)
Gave(Max, Max, Carl)
Gave(Max, Claire, Carl)
Gave(Max, Scruffy, Carl)
Gave(Max, Carl, Carl)
Gave(Claire, Max, Max)
Gave(Claire, Claire, Max)
Gave(Claire, Scruffy, Max)
Gave(Claire, Carl, Max)
Gave(Claire, Max, Claire)
Gave(Claire, Claire, Claire)
Gave(Claire, Scruffy, Claire)
Gave(Claire, Carl, Claire)
Gave(Claire, Max, Scruffy)
Gave(Claire, Claire, Scruffy)
Gave(Claire, Scruffy, Scruffy)
Gave(Claire, Carl, Scruffy)
Gave(Claire, Max, Carl)
Gave(Claire, Claire, Carl)
Gave(Claire, Scruffy, Carl)
Gave(Claire, Carl, Carl)
Gave(Scruffy, Max, Max)
Gave(Scruffy, Claire, Max)
Gave(Scruffy, Scruffy, Max)
Gave(Scruffy, Carl, Max)
Gave(Scruffy, Max, Claire)
Gave(Scruffy, Claire, Claire)
Gave(Scruffy, Scruffy, Claire)
Gave(Scruffy, Carl, Claire)
Gave(Scruffy, Max, Scruffy)
Gave(Scruffy, Claire, Scruffy)
Gave(Scruffy, Scruffy, Scruffy)
Gave(Scruffy, Carl, Scruffy)
Gave(Scruffy, Max, Carl)
Gave(Scruffy, Claire, Carl)
Gave(Scruffy, Scruffy, Carl)
Gave(Scruffy, Carl, Carl)
Gave(Carl, Max, Max)
Gave(Carl, Claire, Max)
Gave(Carl, Scruffy, Max)
Gave(Carl, Carl, Max)
Gave(Carl, Max, Claire)
Gave(Carl, Claire, Claire)
Gave(Carl, Scruffy, Claire)
Gave(Carl, Carl, Claire)
Gave(Carl, Max, Scruffy)
Gave(Carl, Claire, Scruffy)
Gave(Carl, Scruffy, Scruffy)
Gave(Carl, Carl, Scruffy)
Gave(Carl, Max, Carl)
Gave(Carl, Claire, Carl)
Gave(Carl, Scruffy, Carl)
Gave(Carl, Carl, Carl)
Thus, there are 64 possible sentences. Perhaps the number of expressions is given by na where n is the number of
names and a is the arity.
The first language would need the same fours names (Max, Claire, Scruffy, and Carl) and four predicates
(GaveScruffy(x, y), GaveCarl(x, y), GaveMax(x, y), and GaveClaire(x, y)) in order to be able to express everything
that can be said in the second language.
70
Problem 2-15:
1. Claire owned Folly at 2 p.m.
2. Claire gave Silly to Max at 2:05 p.m.
3. Max is a student.
4. Claire erased Folly at 2 p.m.
5. Folly belonged to Max at 3:05 p.m.
6. 2:00 p.m. is earlier than 2:05 p.m.
Owned(Claire, Folly, 2:00)
Gave(Claire, Silly, Max, 2:05)
Student(Max)
Erased(Claire, Folly, 2:00)
Owned(Max, Folly, 3:05)
2:00 < 2:05
Problem 2-16:
1. Owned(Max, Silly, 2:00)
2. Erased(Max, Silly, 2:30)
3. Gave(Max, Silly, Claire, 2:00)
4. (2:00 < 2:00)
Max owned Silly at 2 p.m.
Max erased Silly at 2:30 p.m.
Max gave Silly to Claire at 2 p.m.
2 p.m. is earlier than 2 p.m.
Problem 2-17:
Names
Predicates
Functions
1.
2.
3.
4.
5.
English
AIDS, influenza
Spain, France, Portugal
Misery, Company
Max, Claire, John, Nancy
Jon, Mary Ellen
x is less contagious than y
x is between y and z in size
x loves y
x shook y
x is younger than y
the father of x
the eldest child of x and y
x’s hand
FOL
AIDS, influenza
Spain, France, Portugal
Misery, Company
Max, Claire, John, Nancy
Jon, Mary Ellen
x<y
Between(x, y, z)
Loves(x, y)
Shook(x, y)
Younger(x, y)
father(x)
eldestChild(x, y)
hand(x)
AIDS < influenza
Between(Spain, France, Portugal)
Loves(Misery, Company)
Shook(Max, hand(father(Claire)))
Younger(eldestChild(John, Nancy), eldestChild(Jon, Mary Ellen))
Problem 2-18:
Proof: Suppose that a = b and b = c (two given premises). According to Ind Id, we can replace b in a = b by c. We come
up with a = c, as desired.
Problem 2-19:
1. YES – if a is to the left of b, it follows that b is to the right of a
2. YES – if a is to the left of b, and b =c, one can say a is to the left of c. It follows that c is to the right of a.
3. NO – no details given about the relationship of b and c – see 2-19-3.wld:
OR
4. YES – from the first statement, a front-to-back order is given as (b, a). The second statement adds that a is in front of
c, so c must be placed after a in the list: (b, a, c). The conclusion holds true in this ordering.
5. NO – see 2-19-5.wld:
71
Problem 2-20:
1. Owned(Claire, Folly, 2:00)
2. Owned(Max, Silly, 2:00)
3. ¬Owned(Claire, Silly, 2:00)
- If it is assumed that a disk can only have one owner at a given time, (3) follows from (1) and (2).
- (2) does not follow from (1) and (3). These statements do not say anything about Silly’s owner at 2 p.m. other than
that it was not Claire. For example, let Silly belong to Mary at 2 pm. So, under this situation, (1) is true, (3) is true,
but (2) is false.
- (1) does not follow from (2) and (3). There is nothing stated about Folly and nothing that prevents Max from
owning two disks, as in Owned(Max, Folly, 2:00). For example, let Folly and Silly belong to Max at 2 p.m. Then
(2) is true and (3) is true, but (1) is false in this case.
Problem 2-21:
1. a = b
2. b = c
3. a = c
Ind Id: 1, 2
Problem 2-22:
1. Likes(a, b)
2. b = c
3. c = d
4. b = d
5. Likes(a, d)
Ind Id: 2, 3
Ind Id: 1, 4
Problem 2-23:
1. Between(a, d, b)
2. a = c
3. e = b
4. Between(c, d, b)
5. e = e
6. b = e
7. Between(c, d, e)
Ind Id: 1, 2
Refl =
Ind Id: 5, 3
Ind Id: 4, 6
(using property 1, replacing a with c – replace left of = with right of =)
(no e in 1 or 4 on the left, want it on the left to replace)
(using property 4, replacing b (left) with e(right) from 6)
Problem 2-24:
First, set up a rule: Transitivity of Smaller (Trans. Sm)
#m Smaller (a, b)
:
#n Smaller (b, c)
:
Smaller (a, c)
Trans. Sm: #m, #n
Then, the proof is:
1. Smaller(a, b)
2. Smaller(c, d)
3. b = c
4. Smaller(a, c)
Ind Id: 1, 3
5. Smaller(a, d)
Trans. Sm: 4, 2
Problem 2-25:
First, set up a rule: Left-Right Symmetry (L-R Sym)
#m RightOf(a, b)
:
:
LeftOf(b, a)
L-R Sym: m
The proof is:
1. LeftOf(b, a)
2. b = c
3. LeftOf(c, a)
4. RightOf(a, c)
Ind Id: 1, 2
L-R Sym: 3
72
Problem 3-1: 3-1.sen (containing only ¬¬¬¬¬Between(c, b, d)) and wittgens.wld; Tarski’s World Drill
The count of negation symbols is odd, so the atomic sentence is negated. Since the atomic sentence is true, this claim is
false. As playing of the game proceeds, the number of negation symbols is decremented at each step, and the
commitment switches from a true statement to a false statement.
Problem 3-3: 3-3.sen (as presented in the text) and wittgens.wld; Tarski’s World Drill
Shape of f
Size of f
Number of true sentences in wittgens.wld
Tet
Small
5
Tet
Medium
6
Tet
Large
5
Cube
Small
4
Cube
Medium
4
Cube
Large
4
Dodec
Small
4
Dodec
Medium
4
Dodec
Large
4
At most 6 sentences will be true at once. It would not be possible for all to be simultaneously true, since some pairs,
such as 1 & 2 contradict each other. The minimum number of sentences that will be true is 4.
Problem 3-4: 3-4.sen (modification of sentences provided in the text) and wittgens.wld; Tarski’s World Drill
Shape of f
Size of f
Number of true sentences in wittgens.wld
Tet
Small
6
Tet
Medium
6
Tet
Large
6
Cube
Small
5
Cube
Medium
6
Cube
Large
5
Dodec
Small
5
Dodec
Medium
6
Dodec
Large
5
Again, at most 6 sentences will be simultaneously true. However, the minimum number of sentences that will be true is
5 instead of 4 since only one of two disjuncts must be true but both conjuncts must be true in order to make the
corresponding compound sentence true.
Problem 3-5: bool.sen and wittgens.wld; Yes; Tarski’s World Drill
Problem 3-6: schrder.sen and 3-6.wld:
L
M
M
M
M
S
73
Problem 3-7: demorg.sen and 3-7-1.wld and 3-7-2.wld; For each odd numbered sentence n, sentence n+1 will have the
same truth value because each even numbered sentence is logically equivalent to the odd numbered sentence that directly
precedes it due to DeMorgan’s Laws;
3-7-1.wld:
3-7-2.wld:
S
S
S
M
S
Problem 3-9:
1. (A ∧ B) ∧ A
A∧B∧A
A∧A∧B
A∧B
⇔ (B ∧ A ∧ B ∧ C)
⇔ B∧A∧B∧C
⇔ A∧B∧B∧C
⇔ A ∧ B ∧ C
3. (A ∨ B) ∨ (C ∧ D) ∨ A ⇔ (A ∨ B) ∨ (C ∧ D) ∨ A
⇔ A ∨ (A ∨ B) ∨ (C ∧ D)
⇔ A ∨ A ∨ B ∨ (C ∧ D)
⇔ A ∨ B ∨ (C ∧ D)
4. (¬A ∨ B) ∨ (B ∨ C) ⇔ ¬A ∨ B ∨ B ∨ C
⇔ ¬A ∨ B ∨ C
5. (A ∧ B) ∨ C ∨ (B ∧ A) ∨ A ⇔ (A ∧ B) ∨ (B ∧ A) ∨ C ∨ A
⇔ (A ∧ B) ∨ (A ∧ B) ∨ C ∨ A
⇔ (A ∧ B) ∨ C ∨ A
⇔ A ∨ (A ∧ B) ∨ C
⇔
⇔
⇔
2. (B ∧ (A ∧ B ∧ C))
S
74
Problem 3-10:
1. ¬(Home(Carl) ∧ ¬Home(Claire)) ⇔ ¬Home(Carl) ∨ ¬¬Home(Claire)
⇔ ¬Home(Carl) ∨ Home(Claire)
2. ¬[Happy(Max) ∧ (¬Likes(Carl, Claire) ∨ ¬Likes(Claire, Carl))
⇔ ¬Happy(Max) ∨ ¬(¬Likes(Carl, Claire) ∨ ¬Likes(Claire, Carl))
⇔ ¬Happy(Max) ∨ (¬¬Likes(Carl, Claire) ∧ ¬¬Likes(Claire, Carl))
⇔ ¬Happy(Max) ∨ (Likes(Carl, Claire) ∧ Likes(Claire, Carl))
3. ¬¬¬[(Happy(Max) ∨ Home(Carl)) ∧ (Happy(Max) ∨ Happy(Carl))]
⇔ ¬[(Happy(Max) ∨ Home(Carl)) ∧ (Happy(Max) ∨ Happy(Carl))]
⇔ ¬(Happy(Max) ∨ Home(Carl)) ∨ ¬ (Happy(Max) ∨ Happy(Carl))
⇔ (¬Happy(Max) ∧ ¬Home(Carl)) ∨ (¬Happy(Max) ∧ ¬Happy(Carl))
Problem 3-12: 3-12.sen
Problem 3-15:
1. Max is a student, not a disk.
Student(Max) ∧ ¬Disk(Max)
2. Claire erased Folly at 2 p.m. and then gave it to Max.
Erased(Claire, Folly, 2:00) ∧ Gave(Claire, Folly, Max, 2:00)
3. Folly belonged to either Max or Claire at 2:05 p.m.
Owned(Max, Folly, 2:05) ∨ Owned(Claire, Folly, 2:05)
4. Neither Max nor Claire erased Folly at 2 p.m. or at 2:05 p.m.
¬[Erased(Max, Folly, 2:00) ∨ Erased(Claire, Folly, 2:00)] ∧ ¬[Erased(Max, Folly, 2:05) ∨ Erased(Claire,Folly,2:05)]
Problem 3-15: continued
5. 2:00 p.m. is between 1:55 p.m. and 2:05 p.m.
(1:55 < 2:00) ∧ (2:00 < 2:05)
6. When Max gave Folly to Claire at 2 p.m., 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. Claire is a student, but Max is not.
2. Silly is a disk, and Max did not own it at 2 p.m. OR Silly is a disk that Max did not own at 2 p.m.
3. Claire owned Silly or Folly at 2 p.m.
4. Max didn’t erase both Silly and Folly at 2 p.m.
5. When Max gave either Silly or Folly to Claire at 2 p.m., it was blank, but Claire was angry five minutes later.
Problem 3-17:
English
FOL
AIDS, influenza, Abe, Stephen,
AIDS, influenza, Abe, Stephen,
Sunday, Monday, Dan, George, Al, Sunday, Monday, Dan, George, Al,
Names
Bill, Daisy, Dee, Polonius
Bill, Daisy, Dee, Polonius
x is less contagious than y
LessContagious(x, y)
x is more deadly than y
MoreDeadly(x, y)
x is less deadly than y
LessDeadly(x, y)
x fooled y on day d
Fooled(x, y, d)
x admires y
Admires(x, y)
x is a miller
Miller(x)
Predicates
x is jolly
Jolly(x)
x lives near y
LivesNear(x, y)
x is a borrower
Borrower(x)
x is a lender
Lender(x)
x is a river
River(x)
the eldest child of x
eldestChild(x)
Functions
1. LessContagious(AIDS, influenza) ∧ MoreDeadly(AIDS, influenza)
2. Fooled(Abe, Stephen, Sunday) ∧ ¬Fooled(Abe, Stephen, Monday)
3. (Admires(Dan, Al) ∧ Admires(Dan, Bill)) ∨ (Admires(George, Al) ∧ Admires(George, Bill))
4. Miller(Daisy) ∧ Jolly(Daisy) ∧ River(Dee) ∧ LivesNear(Daisy, Dee)
5. ¬(Borrower(eldestChild(Polonius)) ∨ Lender(eldestChild(Polonius)))
75
76
Problem 3-39: { P ∨ (Q ∧ R) } ╞ (P ∨ Q) ∧ (P ∧ R)
1. P ∨ (Q ∧ R)
2. P
3. P ∨ Q
∨ Intro: 2
∨ Intro: 2
4. P ∨ R
∧ Intro: 3, 4
5. (P ∨ Q) ∧ (P ∨ R)
6. Q ∧ R
∧ Elim: 6
7. Q
∨ Intro: 7
8. P ∨ Q
9. R
∧ Elim: 6
∨ Intro: 9
10. P ∨ R
∧ Intro: 8, 10
11. (P ∨ Q) ∧ (P ∨ R)
12. (P ∨ Q) ∧ (P ∨ R)
∨ Elim: 1, 2-5, 6-11
Problem 3-40:
1. { P ∧ Q } ╞ P ∨ Q
1. P ∧ Q
2. P
3. P ∨ Q
∧ Elim: 1
∨ Intro: 2
2. { (a = b) ∧ (b = c) } ╞ a = c
1. (a = b) ∧ (b = c)
2. a = b
3. b = c
4. a = c
∧ Elim: 1
∧ Elim: 1
Ind Id: 2, 3
3. { (A ∧ B) ∨ C } ╞ C ∨ B
1. (A ∧ B) ∨ C
2. A ∧ B
3. B
4. C ∨ B
5. C
6. C ∨ B
7. C ∨ B
∧ Elim: 2
∨ Intro: 3
∨ Intro: 5
∨ Elim: 1, 2-4, 5-6
4. { A, B ∨ C } ╞ (A ∧ B) ∨ (A ∧ C)
1. A
2. B ∨ C
3. B
4. A ∧ B
∧ Intro: 1, 3
∨ Intro: 4
5. (A ∧ B) ∨ (A ∧ C)
6. C
7. A ∧ C
∧ Intro: 1, 7
∨ Intro: 7
8. (A ∧ B) ∨ (A ∧ C)
∨ Elim: 1, 3-5, 6-8
9. (A ∧ B) ∨ (A ∧ C)
Problem 3-41: { P ∨ Q, ¬ P } ╞
1. P ∨ Q
2. ¬ P
3. P
4. ¬ Q
5. P ∧ ¬ P
6. ¬¬ Q
7. Q
8. Q
9: Q
10. Q
Q
∧ Intro: 3, 2
¬ Intro: 4-5
¬ Elim: 6
Reit: 8
∨ Elim: 1, 3-7, 8-9
Problem 3-42: ∅ ╞ P ∨ ¬ P
1. ¬(P ∨ ¬ P)
2. P
3. P ∨ ¬P
∨ Intro: 2
4. (P ∨ ¬ P) ∧ ¬(P ∨ ¬ P)
∧ Intro: 3, 1
5. ¬P
6. ¬P
∨ Intro: 6
7. P ∨ ¬P
8. (P ∨ ¬ P) ∧ ¬(P ∨ ¬ P)
∧ Intro: 7, 1
9. ¬¬P
¬ Intro: 6-8
∧ Intro: 5, 9
10. ¬P ∧ ¬¬P
11. ¬¬(P ∨ ¬P)
¬ Intro: 1-10
12. P ∨ ¬P
¬ Elim: 11
Problem 3-43: { P ∨ Q, P ∨ R } ╞ P ∨ (Q ∧ R)
1. P ∨ Q
2. P ∨ R
3. P
∨ Intro: 3
4. P ∨ (Q ∧ R)
5. Q
6. P
∨ Intro: 6
7. P ∨ (Q ∧ R)
8. R
9. Q ∧ R
∧ Intro: 5, 8
∨ Intro: 9
10. P ∨ (Q ∧ R)
∨ Elim: 2, 6-7, 8-10
11. P ∨ (Q ∧ R)
12. P ∨ (Q ∧ R)
∨ Elim: 1, 3-4, 5-11
Problem 3-45: { Cube(c) ∨ Dodec(c), Tet(b) } ╞
¬(b=c)
1. Cube(c) ∨ Dodec(c)
2. Tet(b)
3. b = c
4. Tet(c)
Ind Id: 2, 3
5. Tet(c) ∧ (Cube(c) ∨ Dodec(c))* ∧ Intro: 4, 1
6. ¬(b = c)
¬ Intro: 3-5
* Tet(c) ∧ (Cube(c) ∨ Dodec(c)) is a contradiction since
a block cannot be both a tet. and cube (or a dodec.).
Problem 3-46:
1. { ¬(P ∨ Q) } ╞ ¬P ∧ ¬Q
1. ¬(P ∨ Q)
2. P
3. P ∨ Q
4. (P ∨ Q) ∧ ¬(P ∨ Q)
5. ¬P
6. Q
7. P ∨ Q
8. (P ∨ Q) ∧ ¬(P ∨ Q)
9. ¬Q
10. ¬P ∧ ¬Q
∨ Intro: 2
∧ Intro: 3, 1
¬ Intro: 2-4
∨ Intro: 6
∧ Intro: 3, 1
¬ Intro: 6-8
∧ Intro: 5, 9
2. { ¬P ∧ ¬Q } ╞ ¬(P ∨ Q)
1. ¬P ∧ ¬Q
2. P ∨ Q
3. P
4. ¬P
5. P ∧ ¬P
6. Q
7. ¬Q
8. ¬(P ∧ ¬P)
9. Q ∧ ¬Q
10. ¬¬(P ∧ ¬P)
11. P ∧ ¬P
12. P ∧ ¬P
13. ¬(P ∨ Q)
Problem 3-47:
1. { ¬(A ∨ B) } ╞ ¬A
1. ¬(A ∨ B)
2. A
3. A ∨ B
4. (A ∨ B) ∧ ¬(A ∨ B)
5. ¬A
77
∧ Elim: 1
∧ Intro: 3, 4
∧ Elim: 1
∧ Intro: 6, 8
¬ Intro: 8-9
¬ Elim: 10
∨ Elim: 2, 3-5, 6-11
¬ Intro: 2-12
B
T
F
T
F
1. a = b ∧ b ≠ a
2. a = b
3. b ≠ a
4. b ≠ b
5. ¬(a = b ∧ b ≠ a)
∨ Intro: 2
∧ Intro: 3, 1
¬ Intro: 2-4
(A → B) ∧ (B → A)
T
T
T
F
T
F
T
F
F
T
T
T
∨ Intro: 4
∧ Intro: 5, 2
¬ Intro: 4-6
¬ Elim: 7
∧ Intro: 3, 8
∧ Intro: 9, 1
¬ Intro: 11
¬ Elim: 11
3. ∅ ╞ ¬(a = b ∧ b ≠ a)
Problem 4-1:
A
T
T
F
F
2. { ¬(¬A ∧ B), ¬(¬B ∨ C) } ╞ A
1. ¬(¬A ∧ B)
2. ¬(¬B ∨ C)
3. ¬A
4. ¬B
5. ¬B ∨ C
6. (¬B ∨ C) ∧ ¬(¬B ∨ C)
7. ¬¬B
8. B
9. ¬A ∧ B
10. (¬A ∧ B) ∧ ¬(¬A ∧ B)
11. ¬¬A
12. A
A↔B
T
F
F
T
∧ Elim: 1
∧ Elim: 1
Ind Id: 3, 2
¬ Intro: 1-4
78
Problem 4-9:
1. Gave(Claire, Folly, Max, 2:03) Æ (Owned(Claire, Folly, 2:00) ∧ Owned(Max, Folly, 2:05))
2. Erased(Max, Folly, 2:00) ∧ (Gave(Max, Folly, Claire, 2:00) Æ ¬Blank(Folly, 2:05))
3. ¬(Erased(Max, Folly, 2:00) ∨ Erased(Claire, Folly, 2:00)) Æ ¬Blank(Folly, 2:00)
4. Angry(Max, 2:05) Æ (Erased(Claire, Silly, 2:00) ∨ Erased(Claire, Folly, 2:00))
5. Student(Max) ↔ ¬Student(Claire)
Problem 4-10:
1. If Max or Claire erased Folly at 2 p.m., then Folly is a disk.
2. Max erased Folly at 2:30 p.m. if and only if Claire erased Silly at 2 p.m.
3. If Folly wasn’t blank at 2 p.m., Silly was. OR Silly was blank at 2 p.m. unless Folly was.
4. It is not the case that if Folly wasn’t blank at 2 p.m., then Silly was also.
79
80
Problem 4-11:
English
FOL
Abe
Abe
Stephen
Stephen
Ulysses
Ulysses
you
you
me
me
France
France
Treaty
the treaty
Names
Germany
Germany
Tweedledee
Tweedledee
Party
a party
John
John
Mary
Mary
Concert
the concert
x can fool y
CanFool(x, y)
x scratches y
Scratch(x, y)
x will sign y
Sign(x, y)
Predicates
x gets y
Gets(x, y)
x and y went to z together
WentTogether(x, y, z)
x likes y
Like(x, y)
x’s back
back(x)
Functions
1. CanFool(Abe, Stephen) Æ CanFool(Abe, Ulysses)
2. Scratch(you, back(me)) Æ Scratch(me, back(you))
3. Sign(France, Treaty) Æ Sign(Germany, Treaty)
4. Gets(Tweedledee, Party) ↔ Gets(Tweedledum, Party)
5. WentTogether(John, Mary, Concert) Æ (Like(John, Mary) ∧ Like(Mary, John))
A Supplemental Problem: Expressing the basic logic operators using only two (→ and ¬):
The results:
¬P
¬P
EVEN THOUGH the
P∨Q
¬P → Q
language of first order logic
P∧Q
¬(P → ¬Q)
provides four basic logical
symbols (so far), only two of
P→Q
P→Q
them are truly necessary.
This problem shows how to
Truth tables for the ∧ and ∨:
express every type of
P
Q
P ∧ Q ¬ ( P → ¬ Q)
P∨Q
¬ P → Q
sentence using only ¬ and →.
T
T
F T T T
T
T T F F T
T
The same can be done using
(¬ and ∧) or (¬ and ∨).
T
F
F T T F
F
F T T T F
T
F
F
T
F
F
F
F F
F F
T
F
T
T
T
F
T
F
T
T
F
F
T
F
T
F
81
Problem 4-23:
1. Modus Tollens: { A → B, ¬B } ╞ ¬A
1. A → B
2. ¬B
3. A
4. B
→ Elim: 1, 3
5. B ∧ ¬B
∧ Intro: 4, 2
6. ¬A
¬ Intro: 3-5
Problem 4-24:
1. ∅ ╞ A → (B → A)
2. Strengthening the Antecedent:
{ B → C }╞ (A ∧ B)→C
1. B → C
2. A ∧ B
3. B
4. C
5. (A ∧ B) → C
2. ∅ ╞ [A → (B→ C)] ↔ [(A ∧ B) → C]
∧ Elim: 2
→ Elim: 1, 3
→ Intro: 2-4
3. Weakening the Consequent: { A → B }╞ A → (B∨ C)
1. A → B
2. A
3. B
→ Elim: 1, 2
4. B ∨ C
∧ Intro: 3
5. A → (B ∨ C)
→ Intro: 2-4
4. Constructive Dilemma: {A ∨ B, A→C, B→D}╞ C ∨ D
1. A ∨ B
2. A → C
3. B → D
4. A
5. C
→ Elim: 2, 4
6. C ∨ D
∨ Intro: 5
7. B
8. D
→ Elim: 3, 7
9. C ∨ D
∨ Intro: 8
10. C ∨ D
∨ Elim: 1, 4-6, 7-9
5. Trans. of Biconditional: {A ↔ B, B ↔ C } ╞ A ↔ C
1. A ↔ B
2. B ↔ C
3. A
4. B
↔ Elim: 1, 3
5. C
↔ Elim 2, 4
6. C
7. B
↔ Elim: 2, 6
8. A
↔ Elim: 1, 6
9. A ↔ C
↔ Intro: 3-5, 6-8
1. A
2. B
3. A
4. B → A
5. A → (B → A)
Reit: 1
→ Intro: 2-3
→ Intro: 1-4
1. A → (B → C)
2. A ∧ B
3. A
∧ Elim: 2
4. B
∧ Elim: 2
5. B → C
→ Elim: 1, 3
6. C
→ Elim: 5, 4
7. (A ∧ B) → C
→ Intro: 2-6
8. (A ∧ B) → C
9. A
10. B
11. A ∧ B
∧ Intro: 9, 10
12. C
→ Elim: 8, 11
13. B → C
→ Intro: 10-12
14. A → (B → C)
→ Intro: 9-13
15. [A → (B→ C)] ↔ [(A ∧ B) → C]
↔ Intro: 1-7, 8-14
3. { A ∨ (B ∧ C), ¬E, (A ∨ B) → (D ∨ E), ¬A } ╞ C ∧ D
Proof Version 1:
1. A ∨ (B ∧ C)
2. ¬E
3. (A ∨ B) → (D ∨ E)
4. ¬A
5. B ∧ C
6. B
∧ Elim: 5
7. A ∨ B
∨ Intro: 6
8. D ∨ E
→ Elim: 3, 7
9. D
10. C
∧ Elim: 5
11. C ∧ D
∧ Intro: 9, 10
12. E
13. ¬(C ∧ D)
14. E ∧ ¬E
∧ Intro: 12, 2
15. ¬¬(C ∧ D)
¬ Intro: 13-14
16. C ∧ D
¬ Elim: 15
17. C ∧ D
∨ Elim: 8, 9-11, 12-16
18. A
19. ¬(C ∧ D)
20. A ∧ ¬A
∧ Intro: 18, 4
21. ¬¬(C ∧ D)
¬ Intro: 19-20
22. C ∧ D
¬ Elim: 21
23. C ∧ D
∨ Elim: 1, 5-17, 18-22
82
3. {A ∨ (B ∧ C), ¬E, (A ∨ B) → (D ∨ E), ¬A}╞ C ∧ D
Proof Version 2:
1. A ∨ (B ∧ C)
2. ¬E
3. (A ∨ B) → (D ∨ E)
4. ¬A
5. A
6. ¬( B ∧ C)
7. A ∧ ¬A
∧ Intro: 4, 5
8. ¬¬(B ∧ C)
¬ Intro: 6-7
9. B ∧ C
¬ Elim: 8
10. B ∧ C
11. B ∧ C
Reit: 10
12. B ∧ C
∨ Elim: 1, 5-9, 10-11
13. B
∧ Elim: 12
14. A ∨ B
∨ Intro: 13
15. D ∨ E
→ Elim: 3, 14
16. D
17. D
Reit: 16
18. E
19. ¬D
20. E ∧ ¬E
∧ Intro: 18, 2
21. ¬¬D
¬ Intro: 21
22. D
¬ Elim: 22
∨ Elim: 15, 16-17, 18-22
23. D
24. C
∧ Elim: 12
25. C ∧ D
∧ Intro: 24, 23
Problem 4-25:
1. Logical Equiv: (P → Q) ↔ (¬P ∨ Q)
1. P → Q
2. ¬(¬P ∨ Q)
3. ¬P
4. ¬P ∨ Q
5. (¬P ∨ Q) ∧ ¬(¬P ∨ Q)
6. ¬¬ P
7. P
8. Q
9. ¬P ∨ Q
10. (¬P ∨ Q) ∧ ¬(¬P ∨ Q)
11.¬¬(¬P ∨ Q)
12. ¬P ∨ Q
13. ¬P ∨ Q
14. P
15. Q
16. Q
17. ¬P
18. ¬Q
19. P ∧ ¬P
20. ¬¬Q
21. Q
22. Q
23. P → Q
24. (P → Q) ↔ (¬P ∨ Q)
∨ Intro: 3
∧ Intro: 4, 2
¬ Intro: 3-5
¬ Elim: 6
→ Elim: 1, 7
∨ Intro: 8
∧ Intro: 9, 2
¬ Intro: 2-10
¬ Elim: 11
Reit: 15
∧ Intro: 14, 17
¬ Intro: 18-19
¬ Elim: 20
∨ Elim:13,15-16,17-21
→ Intro: 14-22
↔ Intro: 1-12, 13-23
2. Logical Equiv: ¬(P → Q) ↔ (P ∧ ¬Q)
1. ¬(P → Q)
2. ¬(P ∧ ¬Q)
3. ¬(¬P ∨ Q)
4. ¬P
5. ¬P ∨ Q
6. (¬P ∨ Q) ∧ ¬(¬P ∨ Q)
7. ¬¬P
8. P
9. Q
10. ¬P ∨ Q
11. (¬P ∨ Q) ∧ ¬(¬P ∨ Q)
12. ¬Q
13. P ∧ ¬Q
14. (¬P ∧Q) ∧ ¬(¬P∧Q)
15. ¬¬(¬P ∨ Q)
16. ¬P ∨ Q
17. P
18. Q
19. Q
20. ¬P
21. ¬Q
22. P ∧ ¬P
23. ¬¬Q
24. Q
25. Q
26. P → Q
27. (P → Q) ∧ ¬(P → ¬Q)
28. ¬¬(P ∧ ¬Q)
29. P ∧ ¬Q
30. P ∧ ¬Q
31. P → Q
32. P
33. Q
34. ¬Q
35. Q ∧ ¬Q
36. ¬(P → Q)
37. ¬(P → Q) ↔ (P ∧ ¬Q)
∨ Intro: 4
∧ Intro: 5, 3
¬ Intro: 4-6
¬ Elim: 7
∨ Intro: 9
∧ Intro: 10, 3
¬ Intro: 9-11
∧ Intro: 8, 12
∧ Intro: 13, 2
¬ Intro: 3-14
¬ Elim: 15
Reit: 18
∧ Intro: 17, 20
¬ Intro: 21-22
¬ Elim: 23
∨ Elim: 16,18-19,20-24
→ Intro: 17-25
∧ Intro: 26, 1
¬ Intro: 2-27
¬ Elim: 28
∧ Elim: 30
→ Elim: 31, 32
∧ Elim: 30
→ Elim: 33, 34
¬ Intro: 31-35
↔ Intro: 1-29, 30-36
83
84
Problem 5-3: 5-3.sen continued:
Problem 5-6: Reserved for Logical Project One (the solution will be provided later)
Problem 5-7: Tarski’s World
• “There is a tetrahedron that is large” == Sentence 2 (∃x (Tet(x) ∧ Large(x))
• “There is a cube between a and b” == Sentence 10 (∃x (Cube(x) ∧ Between(x, a, b)))
• Sentence 5 == “Some block either is a non-tetrahedron or is large.”
• Sentence 6 == “Some block either is a non-tetrahedron or is large.”
∃x (Tet(x) → Large(x)) ⇔ ∃x(¬Tet(x) ∨ Large(x)) ∴Sentence 6 expresses the same claim as Sentence 5.
[Note: for any sentence with the format of ∃x(A(x) → B(x)), one should read it as ∃x(¬A(x) ∨ B(x)).
Problem 5-10:
1. ¬∃x (Prime(x) ∧ Even(x))
2. ∀x (Prime(x) → (¬Even(x) ∨ x = 2))
3. ∃x (Prime(x) ∧ Even(x))
4. ∃x (Prime(x) ∧ ¬Even(x))
F
T
T
T
85
Problem 5-12: Tarski’s World
1. Sentence 1 is the correct translation for “Some cube is large.” Sentence 4 is the correct translation of “All tetrahedral
are small.”
2. 5-12-2.wld: (Note that the single cube can be any size.)
3. No – everything in the world has to be a small tetrahedron, so it’s not possible to create a world where this case is true
but there’s exists a tetrahedron that is not small.
4. No – Sentence 1 requires that there be a large dodecahedron. Sentence 2 will always be true in such a world since
there is a large object. It will also be true in a world where there’s a large object (other than a dodecahedron) or where
there is any object other than a dodecahedron; since ∃x (Dodec(x) → Large(x)) ⇔ ∃x (¬Dodec(x) ∨ Large(x)).
Problem 5-16:
1. ∀x (Person(x) → ¬Disk(x))
2. ∀x (Disk(x) → ¬Person(x))
3. ¬ ∃x (Erased(x, Silly, 2:00) ∨ Erased(x, Silly, 2:05))
4. ∃t (Erased(Claire, Silly, t) ∧ (2:00 < t) ∧ (t < 3:00))
5. ∀x (Student(x) → ∃y(Disk(y) ∧ Gave(Claire, y, x, 2:00))
6. ∀x ((Disk(x) ∧ Owned(Claire, x, 2:00)) → Blank(x, 2:00))
7. Angry(Claire, 3:00) ∧ ∀x ((Student(x) ∧ Angry(x, 3:00)) → x = Claire)
8. ¬∃x (Person(x) ∧ Erased(x, Folly, 2:00))
9. ∀x ((Person(x) ∧ Erased(x, Silly, 2:00)) → Angry(x, 2:00))
10. ∀x ((Person(x) ∧ Owned(x, Silly, 2:00)) → Angry(x, 2:05))
Problem 5-17 :
1. No one owned Folly at 2 p.m.
2. No student erased Folly and was angry at 2 p.m.
3. Anyone to whom Max gave Folly at 2 p.m. was angry at 2:05.
4. Claire never gave Folly to Max.
86
Problem 5-18:
Names
Predicates
Functions
English
I
logic
company
gold
you (should be “one”)
Denmark
x is brave
x knows how to forgive
x is a man
x is a person
x is an island
x cares for y
x is a nation
x deserves y
x is a certainty
x is not y
x is y
x is miserable
x loves y
x glitters
x is a miller
x is jolly
x lived near y
x is a river
x praises y
x is rotten
x is in y
the government of x
the state of x
FOL
I
Logic
Company
Gold
One
Denmark
Brave(x)
Forgive(x)
Man(x)
Person(x)
Island(x)
Care(x, y)
Nation(x)
Deserves(x, y)
Certainty(x)
x ≠ y
x=y
Miserable(x)
Loves(x, y)
Glitters(x)
Miller(x)
Jolly(x)
LivedNear(x, y)
River(x)
Praise(x, y)
Rotten(x)
In(x, y)
govt(x)
state(x)
1. ∀x (Forgive(x) → Brave(x))
2. ∀x (Man(x) → ¬Island(x))
3. ¬∃x Care(x, I) → ¬∃y Care(I, y)
4. ∀x (Nation(x) → Deserves(x, govt(x))
5. Certainty(Logic) ∧ ∀x (x ≠ Logic → ¬Certainty(x))
6. ∀x ((Person(x) ∧ Miserable(x)) → Loves(x, Company))
7. ¬∀x (Glitters(x) → Gold(x))
or ∀x(Glitters(x) → ¬Gold(x))
(These two sentences are not logically equivalent due to the ambiguity of the sentence.)
8. ∃x (Miller(x) ∧ Jolly(x) ∧ River(Dee) ∧ LivedNear(x, Dee))
9. ∀x((Person(x) ∧ ∀y (Person(y) → Praise(x, y)) → ∀y(Person(y) → ¬Praise(x, y))
10. ∃x (In(x, state(Denmark)) ∧ Rotten(x))
Problem 5-19:
• No Ps are Qs == ¬∃x (P(x) ∧ Q(x)) or ∀x (P(x) → ¬Q(x))
• Some Ps are Qs == ∃x (P(x) ∧ Q(x))
• ¬(¬∃x (P(x) ∧ Q(x))) ⇔ ∃x (P(x) ∧ Q(x))
• ¬∀x(P(x) → ¬Q(x)) ⇔ ∃x ¬(P(x) → ¬Q(x))
⇔ ∃x ¬(¬P(x) ∨ ¬Q(x))
⇔ ∃x (¬¬P(x) ∧ ¬¬Q(x))
⇔ ∃x (P(x) ∧ Q(x))
¬ Elim
DeMorgan ∀x/∃x Law
→ Elim
DeMorgan
¬ Elim
87
Problem 5-20:
2 ⇔ 4: ¬∃y (Cube(y) ∧ Large(y)) ⇔ ∀y ¬(Cube(y) ∧ Large(y))
⇔ ∀y (¬Cube(y) ∨ ¬Large(y))
⇔ ∀x (¬Cube(x) ∨ ¬Large(x))
1 ⇔ 5: ¬∀x (Cube(x) → Small(x)) ⇔ ∃x ¬(Cube(x) → Small(x))
⇔ ∃x ¬(¬Cube(x) ∨ Small(x))
⇔ ∃x (¬¬ Cube(x) ∧ ¬ Small(x))
⇔ ∃x (Cube(x) ∧ ¬Small(x))
⇔ ∃u (Cube(u) ∧ ¬Small(u))
⇔ ∃u (¬Small(u) ∧ Cube(u))
3 ⇔ 6: ¬∀x (Large(x) ↔ Dodec(x)) ⇔ ¬∀x ((Large(x) → Dodec(x)) ∧ (Dodec(x) → Large(x)))
⇔ ∃x ¬((Large(x) → Dodec(x)) ∧ (Dodec(x) → Large(x)))
⇔ ∃x ¬((¬Large(x) ∨ Dodec(x)) ∧ (¬Dodec(x) ∨ Large(x)))
⇔ ∃x (¬(¬Large(x) ∨ Dodec(x)) ∨ ¬(¬Dodec(x) ∨ Large(x)))
⇔ ∃x ((¬¬Large(x) ∧ ¬Dodec(x)) ∨ (¬¬Dodec(x) ∧ ¬Large(x)))
⇔ ∃x ((Large(x) ∧ ¬Dodec(x)) ∨ (Dodec(x) ∧ ¬Large(x)))
Problem 5-35:
Proof: ¬∀x P(x) ╞ ∃x ¬P(x)
1. ¬∀x P(x)
2. ¬∃x ¬P(x)
c|
3. ¬P(c)
4. ∃x ¬P(x)
∃ Intro: 3
5. ∃x ¬P(x) ∧ ¬∃x ¬P(x)
∧ Intro: 4, 2
6. ¬¬P(c)
¬ Intro: 3-5
7. P(c)
¬ Elim: 6
8. ∀x P(x)
∀ Intro: c|-7
9. ∀x P(x) ∧ ¬∀x P(x)
∧ Intro: 8, 1
10. ¬¬∃x ¬P(x)
¬ Intro: 2-9
11. ∃x ¬P(x)
¬ Elim: 10
Informally: Suppose ¬∃x ¬P(x) and choose an arbitrary element c of the domain. If it can be shown P(c), one can
conclude that ∀x P(x). To establish P(c), first assume ¬P(c). From this statement one can conclude, by existential
generalization, that there does indeed exist an object without property P, or ∃x ¬P(x). However, this statement
contradicts the first supposition, ¬∃x ¬P(x), so one can conclude that the second supposition ¬P(c) is false. Thus,
¬¬P(c), and P(c) by double negation elimination. Since c is arbitrary, from P(c) one can conclude ∀x P(x). However,
this conclusion contradicts the premise ¬∀x P(x), so the first supposition must be false. Thus, ¬¬∃x ¬P(x), or, by
double negation, ∃x ¬P(x), the desired result. ❏
Problem 5-36:
Corrected:
1. ∀x [(B(x) ∨ T(x)) → (M(x) ∧ G(x))]
2. ∀y [(S(y) ∨ M(y)) → T(y)]
3. ∃x S(x)
b|
4. S(b)
5. S(b) ∨ M(b)
∨ Intro: 4
6. (S(b) ∨ M(b)) → T(b)
∀ Elim: 2
7. T(b)
→ Elim: 6, 5
8. (B(b) ∨ T(b)) → (M(b) ∧ G(b)) ∀ Elim: 1
9. B(b) ∨ T(b)
∨ Intro: 7
10. M(b) ∧ G(b)
→ Elim: 8, 9
11. M(b)
∧ Elim: 10
12. M(b) ∧ S(b)
∧ Intro: 11, 4
13. ∃x (M(x) ∧ S(x))
∃ Intro: 12
14. ∃x (M(x) ∧ S(x))
∃ Elim: 3, |b|-13
As presented in text: WRONG!!!
1. ∀x [(B(x) ∨ T(x)) → (M(x) ∧ G(x))]
2. ∀y [(S(y) ∨ M(y)) → T(y)]
3. ∃x S(x)
b|
4. S(b)
5. S(b) ∨ M(b)
∨ Intro: 4
6. (S(b) ∨ M(b)) → T(b)
∀ Elim: 2
7. T(b)
→ Elim: 6, 5
8. (B(b) ∨ T(b)) → (M(b) ∧ G(b)) ∀ Elim: 1
9. B(b) ∨ T(b)
∨ Intro: 7
10. M(b)
unjustified
11. M(b) ∧ S(b)
∧ Intro: 11, 4
12. ∃x (M(x) ∧ S(x))
∃ Intro: 12
13. ∃x (M(x) ∧ S(x))
∃ Elim: 3, |b|-14
88
Problem 5-37:
1. ∀x [(B(x) ∧ T(x)) → M(x)]
2. ∀y [(T(y) ∨ M(y)) → S(y)]
3. ∃x B(x) ∧ ∃x T(x)
4. ∃x T(x)
|b|
5. T(b)
6. T(b) ∨ M(b)
7. [(T(b) ∨ M(b)) → S(b)]
8. S(b)
9. ∃z S(z)
10. ∃z S(z)
Problem 5-38:
1. ∀y [Cube(y) ∨ Dodec(y)]
2. ∀x [Cube(x) → Large(x)]
3. ∃x ¬Large(x)
|b|
4. ¬Large(b)
5. Cube(b) ∨ Dodec(b)
6. Cube(b)
7. Cube(b) → Large(b)
8. Large(b)
9. ¬Dodec(b)
10. Large(b) ∧ ¬Large(b)
11. ¬¬Dodec(b)
12. Dodec(b)
13. Dodec(b)
14. Dodec(b)
15. Dodec(b)
16. ∃x Dodec(x)
17. ∃x Dodec(x)
∧ Elim: 3
∨ Intro: 5
∀ Elim: 2
→ Elim: 7, 6
∃ Intro: 8
∃ Elim: 4, |b|-9
∀ Elim: 1
∀ Elim: 2
→ Elim: 7, 6
∧ Intro: 8, 4
¬ Intro: 9-10
¬ Elim: 11
Reit: 13
∨ Elim: 5, 6-12, 13-14
∃ Intro: 15
∃ Elim: 3, |b|-16
Problem 5-39:
1. { ∃x(Cube(x) ∧ Small(x)) }╞ ∃x Cube(x) ∧ ∃x Small(x)
1. ∃x(Cube(x) ∧ Small(x))
|b|
2. Cube(b) ∧ Small(b)
3. Cube(b)
∧ Elim: 2
4. ∃x Cube(x)
∃ Intro: 3
5. Small(b)
∧ Elim: 2
6. ∃x Small(x)
∃ Intro: 5
7. ∃x Cube(x) ∧ ∃x Small(x) ∧ Intro: 4, 6
8. ∃x Cube(x) ∧ ∃x Small(x) ∃ Elim: 1, |b|-7
2. The inference { ∃x Cube(x) ∧ ∃x Small(x) } ╞ ∃x (Cube(x) ∧ Small(x)) is invalid → see 5-39-2.wld:
L
S
Based on this world, the premise is true, but the conclusion is false.
89
3. { ∃x Cube(x) ∧ Small(d) } ╞ ∃x (Cube(x) ∧ Small(d))
1. ∃x Cube(x) ∧ Small(d)
2. ∃x Cube(x)
∧ Elim: 1
3. Small(d)
∧ Elim: 1
|b|
4. Cube(b)
5. Cube(b) ∧ Small(d)
∧ Intro: 4, 3
6. ∃x (Cube(x) ∧ Small(d)) ∃ Intro: 5
7. ∃x (Cube(x) ∧ Small(d))
∃ Elim: 2, |b|-6
4. The inference { ∀x (Cube(x) ∨ Small(x)) } ╞ ∀x Cube(x) ∨ ∀x Small(x) is invalid – see 5-39-4.wld:
L
S
S
S
In this world, the premise (All blocks are either cubes or small.) is true, but the conclusion (Either all blocks are
cubes or all blocks are small.) is false.
5. { ∀x Cube(x) ∨ ∀x Small(x) } ╞ ∀x (Cube(x) ∨ Small(x))
1. ∀x Cube(x) ∨ ∀x Small(x)
2. ∀x Cube(x)
|b|
3. Cube(b)
∀ Elim: 2
4. Cube(b) ∨ Small(b)
∨ Intro: 3
5. ∀x (Cube(x) ∨ Small(x)) ∀ Intro: |b|-4
6. ∀x Small(x)
|c|
7. Small(c)
∀ Elim: 6
8. Cube(c) ∨ Small(c)
∨ Intro: 7
9. ∀x (Cube(x) ∨ Small(x)) ∀ Intro: |c|-8
10. ∀x (Cube(x) ∨ Small(x)) ∨ Elim: 1, 2-5, 6-9
Problem 5-40:
1. { ∀y P(y) } ╞ ∀x P(x)
1. ∀y P(y)
|b|
2. P(b)
3. ∀x P(x)
2. { ∃y P(y) } ╞ ∃x P(x)
1. ∃y P(y)
|b|
2. P(b)
3. ∃x P(x)
4. ∃x P(x)
∀ Elim: 1
∀ Intro: |b|-2
∃ Intro: 2
∃ Elim: 1, |b|-3
90
Problem 5-41:
1. { ∃x ¬P(x) } ╞ ¬∀x P(x)
1. ∃x ¬P(x)
|b|
2. ¬P(b)
3. ∀x P(x)
4. P(b)
5. P(b) ∧ ¬P(b)
6. ¬∀x P(x)
7. ¬∀x P(x)
∀ Elim: 3
∧ Intro: 4, 2
¬ Intro: 3-5
∃ Elim: 1, |b|-6
2. { ∀x ¬P(x) } ╞ ¬∃x P(x)
1. ∀x ¬P(x)
2. ∃x P(x)
|b|
3. P(b)
4. ¬P(b)
5. P(b) ∧ ¬P(b)
6. ∃x (P(x) ∧ ¬P(x))
7. ∃x (P(x) ∧ ¬P(x))
8. ¬∃x P(x)
∀ Elim: 1
∧ Intro: 3, 4
∃ Intro: 6
∃ Elim: 2, |b|-6
¬ Intro: 2-7
3. { ¬∃x P(x) } ╞ ∀x ¬P(x)
1. ¬∃x P(x)
2. ¬∀x ¬P(x)
|b|
3. P(b)
4. ∃x P(x)
5. ∃x P(x) ∧ ¬∃x P(x)
6. ¬P(b)
7. ∀x ¬P(x)
8. ∀x ¬P(x) ∧ ¬∀x ¬P(x)
9. ¬¬∀x ¬P(x)
10. ∀x ¬P(x)
∃ Intro: 3
∧ Intro: 1, 4
¬ Intro: 3-5
∀ Intro: |b|-6
∧ Intro: 7, 2
¬ Intro: 2-8
¬ Elim: 9
Problems 6-1: Tarski’s World Drill
Problem 6-2: 6-2.wld
L
S
M
S
S
L
91
L
S
S
M
L
S
Problem 6-4:
Comments:
• 1-3 cannot be made false by adding objects because they all deal with fixed objects.
• 4-10 cannot be made false by adding objects because they all say that a situation is true in at least one case (only ∃,
no ∀). Adding a false case will not change an already existing true case.
• 11 cannot be made false in the existing world since d is in the back row. If it were moved forward a row, an object
could be placed behind it to make it false.
My solution:
Book’s solution:
92
L
L
M
M
Problem 6-14: Reserved for Logical Project Two (the solution will be provided later)
Problem 6-18: 6-18.sen continued:
93
94
Problem 6-23: 6-23.sen
Problem 6-24:
1. ∀x(Student(x) → ∃y∃z(Student(z) ∧ Disk(y) ∧ ∃t Gave(x, y, z, t)))
2. Student(Claire) ∨ ∃x∃t (Disk(x) ∧ Owned(Claire, x, t))
It is assumed that this sentence was asserted at 2 p.m. on January 2, 1996.
3. ¬∃x∃t (Person(x) ∧ Owned(x, Folly, t) ∧ Owned(x, Silly, t))
4. ¬∃x (Student(x) ∧ ∀y(Disk(y) → ∃t Erased(x, y, t)))
5. ¬∃x∃y (Person(x) ∧ Disk(y) ∧ Owned(x, y, 2:00) ∧ Angry(x, 2:00))
6. ¬∃x∃y∃t (Person(x) ∧ Disk(y) ∧ (t < 12:00) ∧ Gave(x, y, Claire, t))
7. ∀x∀t ((Disk(x) ∧ Gave(Max, x, Claire, t)) → (Owned(Claire, x, t) ∧ ¬Owned(Max, x, t)))
8. ¬∃x∃y∃t (Person(x) ∧ ¬Owned(x, y, t) ∧ ∃z (Person(z) ∧ Give(x, y, z, t)))
9. ∀x∀y∀t∀z ((Disk(x) ∧ Owned(Max, x, t) ∧ Erased(Max, x, t)) →
(Disk(y) ∧ Owned(Claire, y, z) ∧ Erased(Claire, y, z) ∧ t < z))
10. ∃x∃t (Disk(x) ∧ (2:00 < t) ∧ (t < 3:00) ∧ Gave(Max, x, Claire) ∧ Blank(x, t))
11. ¬∃x∃y∃z∃t (Student(x) ∧ Disk(y) ∧ Disk(z) ∧ y ≠ z ∧ Owned(x, y, t) ∧ Owned(x, z, t))
12. ∀x∃t∃t’ ((Student(x) ∧ ∃y∃z∃t (Student(x) ∧ Disk(y) ∧ Disk(z) ∧ y ≠ z ∧ Owned(x, y, t’) ∧ Owned(x, z, t’) ∧ t ≤ t’))
→ ∃y∃z(Disk(y) ∧ Disk(z) ∧ y ≠ z ∧ Owned(Claire, y, t) ∧ Owned(Claire, z, t)))
13. ∀x∀y ((Person(x) ∧ Disk(y) ∧ ∀t Owned(x, y, t)) → ∃z Erased(x, y, z))
14. ∀x∀y∀t ((Person(x) ∧ Disk(y) ∧ Owned(x, y, t)) → ∃z (Owned(x, y, z) ∧ Erased(x, y, z)))
15. ∀x∀t [∃y (Person(y) ∧ Erased(y, x, t)) → (Disk(x) ∧ ¬Blank(x, t))]
Problem 6-25:
1. Every student owns a disk.
2. Some student owns every disk.
3. Claire gave someone everything that Max ever gave to her.
4. Claire owns something that Max gave her earlier.
5. Max gave Claire something in the last five minutes.
6. Max gave something to everyone in the last five minutes.
7. Some student gave everyone something in the last five minutes.
95
Problem 6-26:
Names
Predicates
English
you
I
Dee
morn
night
x is a sucker
x was born at t
x is a place
x goes to y at t
x is a soothsayer
x is a truthsayer
x makes a better living than y
x gave y to z
x is required of z
x does right at t
x is a person
x gratifies y
x astonishes y
x is a miller
x is jolly
x is a river
x lived near y at t
x is less than or equal to y
x worked at t
x sang at t
x is a lark
x is more blithe than y
x is company at t
x should expect y at t
x blushes
x needs to blush
x is an animal
x is man
x can fool y at t
x loves y
FOL
You
I
Dee
Morn
Night
Sucker(x)
Born(x, t)
Place(x)
Go(x, y, t)
Soothsayer(x)
Truthsayer(x)
MakeBetterLiving(x, y)
Gave(x, y, z)
Required(x, y)
DoRight(x, t)
Person(x)
Gratify(x, y)
Astonish(x, y)
Miller(x)
Jolly(x)
River(x)
LivedNear(x, y, t)
x ≤ y
Worked(x, t)
Sang(x, t)
Lark(x)
MoreBlithe(x, y)
Company(x, t)
ShouldExpect(x, y, t)
Blushes(x)
NeedsToBlush(x)
Animal(x)
Man(x)
CanFool(x, y, t)
Love(x, y)
Functions
1. ∀t∃x (Sucker(x) ∧ Born(x, t))
2. ∀x∀t ((Place(x) ∧ Go(You, x, t)) → Go(I, x, t))
3. ∀x∀y ((Soothsayer(x) ∧ Truthsayer(y)) → MakeBetterLiving(x, y))
4. ∀x (¬∃y∃z Gave(z, y, x) → ¬∃z Required(z, x))
5. ∀t DoRight(You, t) → {∃x (Person(x) ∧ Gratify(You, x)) ∧∀x [(Person(x) ∧¬Gratify(You, x)) → Astonish(You, x)]}
6. ∃t∃x {Miller(x) ∧ Jolly(x) ∧ River(Dee) ∧ LivedNear(x, Dee, t) ∧
∀z [((Morn ≤ z) ∧ (z ≤ Night)) → (Worked(x, z) ∧ Sang(x, z))] ∧
∀y [Lark(y) → ¬MoreBlithe(y, x)]}
7. ∀t∀x (Company(x, t) → ShouldExpect(You, x, t))
8. ∀x (((Blushes(x) ∨ NeedsToBlush(x)) ∧ Animal(x)) → Man(x))
9. ∀x (Person(x) → [∀y (Person(y) → ∃t CanFool(x, y, t)) ∧ ∃z (Person(z) ∧ ∀u CanFool(x, z, u)) ∧
¬∀y∀t (Person(y) → CanFool(x, y, t)) ]
10. ∀x [ (Person(x) ∧ ∃y (Person(y) ∧ Love(x, y)) ) → ∀z (Person(z) → Love(z, x)) ]
96
Problem 6-32:
English
FOL
Claire
Claire
Max
Max
Names
Melanie
Melanie
Mary
Mary
x is the mother of y
MotherOf(x, y)
Predicates
x is older than y
Older(x, y)
x’s
mother
mother(x)
Functions
1. Older(mother(Claire), mother(Max))
∃x∃y [MotherOf(x, Claire) ∧ MotherOf(y, Max) ∧ ∀z (MotherOf(z, Claire) → x = z)) ∧
∀w (MotherOf(w, Max) → y=w) ∧ x ≠ y ∧ Older(x, y)]
2. ∀x Older(mother(mother(x)), Melanie)
∀x∃y∃z[MotherOf(y, x) ∧ MotherOf(z, y) ∧ x ≠ y ∧ y ≠ z ∧ x ≠ z ∧
∀w[MotherOf(w, x) → y = w] ∧ ∀w [MotherOf(w, y) → z = w] ∧ Older(z, Melanie)]
3. ∃x Older(Mary, mother(mother(x)))
∃x∃y∃z [MotherOf(x, y) ∧ MotherOf(z, y) ∧ x ≠ y ∧ x ≠ z ∧ y ≠ z ∧
∀w (MotherOf(w, x) → y = w) ∧ ∀w (MotherOf(w, y) → z = w) ∧ Older(Mary, z)]
Problem 6-33:
1. [height(father(Max)) > height(Max)] ∧ ¬[height(father(Max)) > height(father(Claire))]
2. ∃x [height(x) > height(father(Claire))]
3. ∀x∃y [height(x) > height(y)]
4. ¬∃x (height(x) > height(x))
5. ∀x [(height(x) > height(Claire)) → (height(x) > height(Max))]
6. ∀x {(height(Claire) > height(x)) → ∃y[(height(y) > height(x)) ∧ (height(father(Max)) > height(y))]}
Problem 7-2:
1. There is exactly one Tove.
2. There is at most one Tove.
3. There is exactly one Tove.
4. There is at most one Tove.
5. Objects are identical iff they are Toves.
Problem 7-3: Reserved for Logical Project Three (the solution will be provided later)

Symbolic Logic Problems

Transcription

Similar documents

Graduate Degrees Offered - University of Wisconsin

thanks to our sponsors - Elim Christian Services

vacation bible school - Elim Lutheran Church

Wiregrass - East Georgia State College

PARK PARK - Downsview Park

2-21-13 Full Board Minutes FINAL

Summer 2015 - LE Phillips Memorial Public Library

these are some in our guestbook

Dossier Open DeLaVallet Bidiefono

Page 1 Mon Wed Thurs Fridays 7:10 AM at the St. Claire Hospital