Solutions to Homework #9, Problems 1-22 (not 15, 17 or 21)
Problem 1. Construct a truth table for the sentence (A  B)  (B  A). Show that its
final column is the same as that for A  B
(A
T
T
F
F

T
F
T
T
B)
T
F
T
F

T
F
F
T
(B
T
F
T
F

T
T
F
T
A)
T
T
F
F
A
T
T
F
F

T
F
F
T
Problem 4. Translate the following English sentences into FOL.
1. If a is a tetrahedron then it is in front of d.
Tet(a)  FrontOf(a,d)
2. a is to one side or the other of d only if it’s a cube.
(LeftOf(a,d)  RightOf(a,d))  Cube(a)
3. c is between either a and e or a and d.
Between(c, a, e)  Between(c, a, d)
4. c is to the right of a if it (c) is small.
Small(c)  RightOf(c, a)
5. c is to the right of d only if b is to the right of c and left of e.
RightOf(c, d)  (RightOf(b, c)  LeftOf(b, e)
6. If e is a tetrahedron, then it’s to the right of b if and only if it is also in front of b.
Tet(e)  (RightOf(e, b)  FrontOf(e, b))
7. If b isn’t in front of d then neither is it in back of d, provided it’s a cube.
Cube(c)  (FrontOf(b, d)  BackOf(b, d))
8. c is in back of a but in front of e.
BackOf(c, a)  FrontOf(c, e)
B
T
F
T
F
9. e is in front of d unless it (i.e., e) is a large tetrahedron.
(Large(e)  Tet(e))  FrontOf(e, d)
10. At least one of a, c, and e is a cube
Cube(a)  Cube(c)  Cube(e)
11. a is a cube only if b is in front of c.
Cube(a)  FrontOf(b, c)
12. b is larger than both a and e
Larger(b, a)  Larger(b, e)
13. a and e are both larger than c, but neither is large
(Larger(a,c)  Larger(e, c))  (Large(a)  Large(e))
or,
(Larger(a,c)  Larger(e, c))  (Large(a)  Large(e))
14. d is the same size as b
(Small(d)  Small(b))  (Medium(d)  Medium(b))  (Large(d)  Large(b))
15. a is large if and only if it’s a cube.
Large(a)  Cube(a)
16. e is a tetrahedron unless c is a cube.
Cube(c)  Tet(e)
17. If e isn’t a tetrahedron, either b or d is small.
Tet(e)  (Small(b)  Small(d))
18. b or d is a dodecahedron if either a or c is a tetrahedron
(Tet(a)  Tet(c))  (Dodec(b) Dodec(d)
19. d is a dodecahedron just in case b is a cube.
Dodec(d)  Cube(b)
20. b is a dodecahedron just in case c is.
Dodec(b)  Dodec(c)
Problem 9. Using the symbols introduced in Table 1, page 23, translate the following
sentences into first-order logic.
1. If Claire gave Folly to Max at 2:03 then it belonged to her at 2:00 and to him at 2:05.
Gave(Claire, Folly, Max, 2:03)  (Owned(Claire,Folly, 2:00)  Owned(Max, Folly,
2:05)
2. Max erased Folly at 2:00 pm , but if he gave it to Claire then, it was not blank five
minutes later.
(Erased(Max, Folly, 2:00))  (Gave(Max, Folly, Claire, 2:00)  Blank(Folly, 2:05)
3. If neither Max nor Claire erased Folly at 2:00, then it wasn’t blank.
(Erased(Max, Folly, 2:00)  Erased(Claire, Folly, 2:00))  Blank(Folly, 2:00)
or,
(Erased(Max, Folly, 2:00)  Erased(Claire, Folly, 2:00))  Blank(Folly, 2:00)
4. Max was angry at 2:05 only if Claire erased either Folly or Silly five minutes before.
Angry(Max, 2:05)  (Erased(Claire, Folly, 2:00)  Erased(Claire, Silly, 2:00))
5. Max is a student if and only if Claire is not.
Student(Max)  Student(Max)
Problem 10. Using Table 1 on page 23, translate the following into colloquial English.
1. (Erased(Max, Folly, 2:00)  Erased(Claire, Folly, 2:00))  Disk(Folly)
If Max erased Folly at 2:00 pm or Claire erased Folly at 2:00 pm then Folly is a disk.
2. Erased(Max, Folly, 2:30)  Erased(Claire, Silly, 2:00)
Max erased Folly at 2:30 just in case Claire erased Silly at 2:00.
Or,
Max erased Folly at 2:30 if and only if Claire erased Silly at 2:00.
3. Blank(Folly, 2:00)  Blank(Silly, 2:00)
Silly was blank at 2:00 unless Folly was blank at 2:00.
4. (Blank(Folly, 2:00)  Blank(Silly, 2:00))
It’s not the case that Silly was blank at 2:00 unless Folly was blank at 2:00.
Problem 11. Translate the following into FOL as best you can. Explain any predicates
and function symbols you use, and any shortcomings in your first-order translations.
1. If Abe can fool Stephen, surely he can fool Ulysses.
English
Names:
Abe
Stephen
Ulysses
Predicates:
x fools y
FOL
Abe
Stephen
Ulysses
Fools(x, y)
Fools(Abe, Stephen)  Fools(Abe, Stephen)
No limitations.
2. If you scratch my back, I’ll scratch yours.
English
Names:
You
Me
Predicates:
x scratches y
Function Symbols:
x’s back
FOL
You
Me
Scratches(x, y)
BackOf(x)
Scratch(You, BackOf(Me))  Scratch(Me, BackOf(You))
Limitations: “You” and “Me” are funky and imprecise.
3. France will sign the treaty only if Germany does.
English
Names:
France
Germany
Predicates:
x will sign the treaty
FOL
France
Germany
SignsTreaty(x)
SignsTreaty(France)  SignsTreaty(Germany)
4. If Tweedledee gets a party, so will Tweedledum, and vice versa.
English
Names:
Tweedledee
Tweedledum
Predicates:
x gets a party
FOL
Tweedledee
Tweedledum
GetsAParty(x)
GetsAParty(TweedleDee)  GetsAParty(Tweedledum)
5. If John and Mary went to the concert together, they must like each other.
English
Names:
John
Mary
Predicates:
x and y went to the Concert together
x likes y
FOL
John
Mary
AttendedConcert(x, y)
Likes(x, y)
AttendedConcert(John, Mary)  (Likes(John, Mary)  Likes(Mary, John)
Problem 12
“Max managed to get Carl home” merely implicates (does not imply) that it was hard to
get Carl home. This is because if we apply Grice’s test, it is possible to add to the
statement in order to negate the implicature without contradicting the initial statement:
“Max managed to get Carl home, and it really wasn’t difficult to do so.”
Problem 13.
If Max asserts that “We can walk to the movie or we can drive,” his assertion logically
implies that we cannot both walk and drive. This is because, given what it means to go
to the movie – that is, we’re not going to go twice and that one cannot walk while driving
or vice versa – to say that “We can walk to the movie or we can drive, or we can do
both.” is a self contradictory statement.
Problem 16. Give proofs of the following valid steps.
1. Modus Tollens: From A  B and B infer A.
Since A  B means A only if B, and since we’re saying that B is not the case (B), then
it follows that A cannot be the case either, hence A.
Another way:
Let’s suppose that A is the case. Since we are assuming A and know A  B, by modus
ponens, this implies B. However, since we also know B, this gives us B and B, a
contradiction. Therefore, by reductio ad absurdum, A.
2. Strenthening the Antecedent: From B  C infer (A  B)  C
In addition to our premise, B  C, let’s suppose A  B. Now, since we are assuming B,
and because we know B  C, C follows, by modus ponens. Since we were able to prove
C by assuming A  B, we know that A  B imply C.
3. Weakening the Consequent: From A  B infer A  (B  C)
In addition to our premise A  B, let’s suppose A. Now, given these statements, B
follow by modus ponens. Since we know B is true, any disjunction containing B will also
be true, therefore we can assert B  C. Since we have derived B  C by assuming A, we
have shown that A  (B  C).
4. Constructive Dilemma: From A  B, A  C, and B  D infer C  D
What I’ll want to show is that C  D follows from our premises no matter which side of
the disjunction A  B is true. Let’s take A first.
Assume A. If A and A  C, C follows by modus ponens. Since we know that C true,
then can assert a disjunction C  D because it doesn’t matter whether D is true or not, by
the definition of .
Now assume B. This, together with B  D imply D, by modus ponens. Since we know
that D is true, then we can assert a disjunction C  D because it doesn’t matter whether C
is true or not, by the definition of .
Since we have shown that C  D follows no matter whether A  B is true and we know
that at least one of them is true, we have shown that our premises imply C  D.
5. Transitivity of the biconditions: From A  B and B  C infer A  C.
A  B states that wherever we have a B, we can replace it with a B. We can therefore
replace the B in B  C with A, yielding A  C.
Problem 18. Does it follow from the following premises that the unicorn is mythical?
That it is magical? Justify your answer.
1. The unicorn, if it is not mythical, is a mammal, but if it is mythical, it is immortal.
2. If the unicorn is either immortal or a mammal, it is horned.
3. The unicorn, if horned, is magical.
It might be easier to do this symbolically:
The unicorn is mythical = Y
The unicorn is a mammal = A
The unicorn is immortal = I
The unicorn is magical = G
The unicorn is horned = H
So, we have
1. (Y  A)  (Y  I)
2. (I  A)  H
3. H  A
Let’s decide whether the unicorn is magical.
Now, the unicorn is either horned or it’s not. If it is horned, it’s magical by 3. But
suppose it isn’t horned if it isn’t horned then (I  H) by modus tollens from 2. By
DeMorgan’s Theorems, this is I A, which would seem to suggest that the unicorn
isn’t magical (A). However, let’s keep on pursuing this line of reasoning. I implies
Y by modus tollens from the second conjunct of 1. However A implies Y by modus
tollens from the first conjunct of 1. This yields Y  Y, a contradiction. Since we
achieved a contradiction by assuming that the unicorn isn’t horned, it must be horned.
Since the unicorn must be magical if it is horned (3), the unicorn must be magical.
It does not seem to follow that it must be mythical (or that it can’t be for that matter).
Problem 20. Give proofs of the following
1. A  (B  A) is a logical truth.
The way we prove that something is a logical truth is by proving it with no premises.
Thus, we need to start out by making assumptions. Let’s assume A. If we can
somehow prove B  A by assuming A, we will have shown that A  (B  A). Now,
how can we prove B  A? Well, we can start out by assuming B. If we can conclude A
(which we can, because we have already assumed it) by assuming B, then we have shown
B  A. Since we have now got B  A by assuming A, we have shown that A  (B 
A) and since we have proven this by starting with no premises, we have shown that it is a
logical truth.
2. (A  (B  C))  ((A  B)  C) is a logical truth.
Since we have a biconditional, we will need to prove the left side by assuming the right
side and the right side by assuming the left side. Let’s assume A  (B  C) and try to
prove ((A  B)  C. Since we need to prove a conditional, let’s assume the antecedent,
namely A  B. If, then, we can prove C based on this assumption, we will have shown
that (A  B)  C. Now that we have assumed A  B, we know A. Since we have
assumed A, it follows that B  C, by modus ponens. Since we have assumed B (this is
implied by A  B), C follows, by modus ponens. Now that we have proven C by
assuming A  B, we have shown that (A  B)  C. Since we got here by assuming A 
(B  C), we have also proven the first half of our biconditional, that (A  (B  C) 
((A  B)  C.
Now we need to show the second half of our biconditional, that we can get (A  (B 
C)) by assuming ((A ^ B)  C. So, let’s assume ((A  B)  C. Since we want to prove
a conditional (A  ( B  C)), we will want to assume the antecedent, A. If we can get
(B  C) by assuming A, we will have shown that A  (B  C). Since we now need to
prove B  C, and this is a conditional, we want to start out by assuming the antecedent
(B). Now that we have assumed (for different reasons) A and B, we know A  B. C
follows from this and our initial assumption (A  B)  C by modus ponens. Since we
have proven C by assuming B, we have shown B  C. Since we got this by assuming A,
we have shown that A  (B  C). Since we got this from our initial assumption of (A 
B)  C, we have shown that ((A  B)  C)  (A  (B  C), the second half of our
biconditional. Since we have shown that the conditional goes both ways and because we
did this with no premises, we have shown that (A  (B  C))  ((A  B)  C) is a
logical truth.
3. A  B is a logical consequence of A  B  C, B  (A  C), and A  C.
Since what we want to show is a biconditional, we need to prove A  B and B  A.
Let’s assume A. Now, since A  C, one of our premises becomes B  (A  A).
However A  A is a contradiction and therefore always false. Since the consequent is
always false, so is the antecedent, by modus tollens, so we have shown that A  B.
Now we need to prove that B  A. Let’s assume B. Since we know that B and we
know that A  B  C, we know that A  C. But since we know that A  C, A and C
must have the same truth value, therefore they are either both true or both false. They
can’t both be false, however, because that would make A  C false. Therefore, A follows
from B. Since we have shown that A  B and B A, we have shown that A 
B.
4. C  D is a consequence of A  (B  C), E, (A  B)  (D  E), and A.
Since we know A and A  (B  C), we know (B  C) from the definition of . Since
we know B  C, we know C. We’re halfway done. Since we know B, we know A  B,
and from this and (A  B)  (D  E), D  E follows by modus ponens. Now, since we
know D  E and E, D follows by the definition of . Since we have shown C and D,
we have also shown C  D.
5. C  B is a consequence of A  B, C  (D  E), D  C, and A  E.
Since we want to prove a conditional, let’s assume the antecedent, C. From C, D  E
follows from C  (D  E) by modus ponens. D also follows given C because of D
 C, by modus tollens. Since we know D and we know D  E, E follows because of
the definition of . A follows from E and A  E by modus tollens. Finally, B
follows from A and from our first premise by modus ponens. Since we have proven B
by assuming C, we have shown that C  B.
Problem 22. Consider the following sentences.
First, we need to translate these sentences.
1. If Max owned Folly at 2:00, then Claire owned Silly at the same time.
1. Owned(Max, Folly, 2:00)  Owned(Claire, Silly, 2:00)
2. Claire owned either Silly or Folly at 2:00
2. Owned(Claire,Silly, 2:00)  Owned(Claire, Folly, 2:00)